Fundamentals of Calculus Concepts

Here are the top 10 most important things to know about calculus. Must know number one, Newton's quotient. To be able to fully understand calculus, you're going to have to understand the logic behind Newton's quotient and how to use it to find the derivative of a function from first principles. Let's do an example where we calculate the derivative from first principles. f of x equals x squared minus 2x plus 4, and then we'll use the derivative to evaluate f prime of 4. Now the derivative of a function is an equation that can calculate the slope or instantaneous rate of change of a function at a single point. I'll use this diagram that I've sketched of this f of x function to help explain this. Let's say for this function f of x that I was interested in finding the slope of it at a single point at some x value. Let's put the point right there. I could label the coordinates of that point, point x f at x. The slope of the function at that single point could be visualized by me drawing a line that is tangent to this function at that point. The slope of this tangent line represents the slope or instantaneous rate of change of f at x at that point. But the challenging part of this is, how do we find the slope of a line where we only know one point that's on the line? Slope is usually calculated using two points and then finding their change in y and then dividing that by their change in their x values. So what we could do is pick another point that's on this f at x function. I'll put the point right here. And let's say that that point is h units to the right of the original point. So its x coordinate would be x plus h and its y coordinate would be the value of the function at x plus h. And now it would be easy to calculate the slope of the line that connects those two points. We can find the slope of that secant line just by doing the change in the y coordinates f at x plus h minus f at x, and then dividing that by the change in the x coordinates x plus h minus x. And those x's would cancel out and in the denominator we'd be left with just an h. Now the slope of that secant line does not match the slope of that tangent line. We could make it a better match or a better approximation by moving that point closer along the function to the original point that we want the slope at. So we can make the h value, which stands for the change in x, we could make it smaller. So let me zoom in to show you how that would work. If I put the point right here and then found this secant line, notice that that secant line more closely matches the slope of the tangent line. So the logical thing to do to get the best approximation for the slope of this tangent line is to move this point. infinitely close to the point we want the slope at. And we have a notation for that. f prime of x. This apostrophe means a lot. It means it's the equation of the derivative, which means it's finding the slope of the tangent at a point. It's the instantaneous rate of change of the function at a point. And to find that, what we want to do is find the limit of the secant slopes, so of that formula, as the change in x, which we call h, approaches zero. So this just communicates that as this point moves infinitely close to the point we want the slope at, the slope of the secant line approaches the slope of the tangent line. And now let's use this formula to find the derivative of that specific f function. So f prime of x equals the limit as h goes to zero of f to evaluate a function at x plus h, I need to sub x plus h in for both of the x's into the function. And then Newton's quotient tells me I then have to subtract f at x. So I need to subtract that function. And that all needs to be divided by h. Now this limit is in indeterminate form right now. We can't do direct substitution yet, but we can expand the numerator, collect our like terms, and get this out of indeterminate form. So I'll do x plus h all squared. I'll distribute the negative 2 to both terms, and I'll subtract all. all three terms of f of x. And now let me zoom in so we can see what we can simplify here. x squared minus x squared, those cancel out. Negative 2x plus 2x equals 0, and 4 minus 4 equals 0. Of all three terms that are left in the numerator, notice they all have an h. So on my next line, I'm going to common factor out that h. And now I know h is actually not going to be equal to 0 because we're finding the limit as h approaches 0. So I know that I can cancel out those h's, and I'll be left with a limit that is equivalent to this limit. And now this limit is out of indeterminate form, so now I can do direct substitution. I can sub 0 in for h, and then I have the equation of my derivative, which of course would simplify to just 2x minus 2. So this equation which we call the equation of the derivative, can find you the slope or instantaneous rate of change of the f at x function at any x value you want. And in this question, it's said to evaluate f prime at 4. That means find the slope of the function at an x value of 4. f prime at 4, we just sub 4 in for x, 2 times 4 minus 2 is 8 minus 2, which is 6. So this means at an x value of 4, the slope. or instantaneous rate of change of the f of x function is 6. Must know number two, derivative rules. Not all derivatives have to be calculated from first principles using Newton's quotient. There are four basic rules which will save you a lot of time. There's the power rule, the product rule, the quotient rule, and the chain rule. Let's start with the power rule. If you're trying to find the derivative, which I could denote as d over dx, of a power function, so x to the n, there's a rule that says you can write the exponent as the coefficient of the power and then reduce the exponent by 1. So the derivative of x to the n with respect to x would be n times x to the n minus 1. And let me do an example to show you how this rule works. Let's find the derivative with respect to x of x to the power of 5. The power rule says take the exponent, write it as the coefficient, and then reduce the value of the exponent by 1. So this becomes 5 times x to the power of 4. Let's try another example. Let's find the derivative with respect to x of 3 times x to the 5. Well when we have a constant factor as part of our derivative, we're actually allowed to move that in front of our derivative operator and then just multiply that by the derivative of the power. And we know the derivative of x to the 5. We write the exponent as the coefficient, keep the base, and reduce the exponent by 1. And then these constant factors of 3 and 5 can be multiplied together to give us 15x to the power of 4. So you can probably see the shortcut that that exponent can just be multiplied by whatever coefficient is already there. 5 times 3 is 15. And then the exponent, 5, reduces by 1 to give you 4. Let's move on to the next rule I want to show you, the product rule. Let's say you're finding the derivative with respect to x of a product of two functions of x. Let's call those functions f at x and g of x. Think of f at x as the first function of the product and g of x as the second function of the product. The product rule says do the derivative of the first function, so f prime of x, multiply that by the second function, plus do the derivative of the second function, and then multiply that by the first function. Let's see how that works. Let's find the derivative with respect to x of x squared plus three times three x minus one. Notice we have a product of two functions of x, x squared plus three is our first function of x three x minus one is our second function of x. If I follow the product rule, it tells me to do the derivative of the first factor. So the derivative of x squared plus three is two x. that gets multiplied by the second factor plus the derivative of the second factor, the derivative of 3x minus 1 is 3, and that gets multiplied by the first factor. So I'll expand this into standard form by distributing the 2x and the 3, and then collecting my like terms gives me 9x squared minus 2x plus 9. The next rule, the quotient rule, which is very related to the product rule, but this time what if I want to find the derivative with respect to x of a quotient of two functions of x? I have a top function and a bottom function. The quotient rule says do the derivative of the top times the bottom function minus the derivative of the bottom function times the top function. And that all has to be divided by the bottom function squared. And then let me make some room here. So let me shrink this rule and we'll do an example. Let's find the derivative of 3x divided by x squared plus 1. Because I have a function of x being divided by another function of x, I need to use the quotient rule, which tells me do the derivative of the top function, the derivative of 3x is 3. That gets multiplied by the bottom function, minus the derivative of the bottom function is 2x, multiplied by the top function. And then don't forget this all gets divided by this bottom function squared. And if I expand the numerator into standard form by distributing this 3, finding this product, and then collecting my like terms, I would get negative 3 x squared. plus 3 divided by that x squared plus 1 squared. And lastly let's learn the chain rule, which is a rule we use when we have composite functions. So let's say we want to find the derivative with respect to x of f of g of x. So it's a function of a function. When you have that, you differentiate the outside function and leave the inside function exactly as it is. but then that has to be multiplied by the derivative of the inside function. So we'd multiply that by g prime of x. Now this chain rule can be applied to lots of different types of functions, but because for these examples we've been focusing on power functions, let me show you how this chain rule specifically applies to power functions. So let's say we're finding the derivative of not just x to the power of n, but f of x to the power of n. Using the power rule, you know we have to write that exponent as the coefficient of the power. We'll leave the base of the power exactly as it is, and then we'll reduce the exponent by one. So we'd have n times f of x to the n minus one. But because the base of the power is a function of x, and it's not just x, we then have to follow a chain rule which says multiply by the derivative of the inside function. So I need to multiply by the derivative of the base of the power. So this would be multiplied by f prime of x. So let's make some room and do an example to show you exactly how this would work. Let's find the derivative with respect to x of 3x minus 2 to the power of 4. This is a power function where the base of the power is a function of x. So following the chain rule, I know I take the exponent, write it as the coefficient of the power. I leave the base of the power exactly as it is. I reduce the exponent by 1. But then I have to remember to multiply this by the derivative. of the inside function. The derivative of 3x minus 2 is 3. And then I could simplify this by multiplying those constant factors together. 4 times 3 is 12. So I have 12 times 3x minus 2 cubed. So let me zoom out here so you can see all four rules that we learned. Must know number three, derivatives of other types of functions. There are three other important types of functions that you need to know how to find the derivative of. Exponential functions, logarithmic functions, and of course trig functions. Let's start with exponential functions. When finding the derivative of an exponential function, so the derivative with respect to x of b to the power of x would be equal to, you keep the power, so b to the power of x stays the exact same, but that gets multiplied by the natural logarithm of the base of the power. So for example, when finding the derivative of five to the power of x, that's just equal to, you keep the power exactly as it is, five to the power of x, and multiply that by the natural logarithm of the base of the power, which is five. And then what if the exponent isn't just x? What if it's a function of x? So let's find the derivative with respect to x of b to the power of f of x. You would need the chain rule. keep the power exactly as it is, still multiply by the natural logarithm of the base, but that would then have to be multiplied by the derivative of the function that's in the exponents. We'd multiply this by f prime of x. Let me make some room for an example, and let's find the derivative with respect to x of 5 to the power of 3x plus 1. Keep the power times the natural logarithm of the base, but then also multiply... by the derivative of the function that's in the exponent. The derivative of 3x plus 1 is 3. And then we'll just simplify that expression by moving those constant factors to the front. Let's move on to logarithmic functions. The derivative of log base b of x is equal to 1 divided by the argument of the logarithm, so x, times the natural logarithm of the base of the logarithm, which is b. An example where we could practice that would be finding the derivative of log base 5 of x. That would be equal to 1 divided by x times the natural logarithm of 5. And then what if the argument wasn't just x? What if it was a function of x? To find the derivative of log base b of f at x, you would need to use the chain rule, which would say follow the same rule 1 divided by the argument which is f times the natural logarithm of the base b, but then you have to multiply all of that by the derivative of the inside function. So multiply by the derivative of f. An example of that would be finding the derivative of log base 5 of 2x cubed. Following the derivative rule, I would do 1 divided by, keep the argument the same, 2x cubed, times the natural logarithm of the base of the log. But then that has to get multiplied by the derivative of the inside function. The derivative of 2x cubed is 6x squared. And then I'll make a little bit of room so I can simplify this. Notice I have a 6 divided by 2, that's 3, and both of those factors of x could cancel with two of those factors of x, leaving me with 3 divided by x times the natural logarithm of 5. And lastly let's move on to trig functions. You need to know how to differentiate the three primary trig functions. The derivative of sin x is equal to cos x, the derivative of cos x is negative sin x, and the derivative of tan x is equal to secant squared of x. And then an example to practice doing derivative rules that involve trig functions, let's find the derivative of 2 times sine of 3x. This constant factor can go in front of the derivative operator, so this would equal 2 times the derivative of sine of 3x. And then when differentiating a sine function, I know the derivative sine of anything is equal to cosine of that same argument. But because the argument of the sine function was not just x, it was a function of x, this is a composite function which means I need chain rule. So I then have to multiply this by the derivative of 3x. The derivative of 3x is 3. And the last thing to do, those constant factors of 2 and 3 can be multiplied together to give 6. So it equals 6 times cosine of 3x. I'll zoom out so you can see all of those rules. Must know number four, the first derivative test. Before I show you the first derivative test, you're going to have to remember what the first derivative tells you about a function. If the value of the first derivative at an x value is greater than zero, so if it's positive, that means that the original function at that x value is negative. increasing. That's because if we have a positive first derivative, that means the slope is positive, which means the function must be going up to the right. If the value of the first derivative at an x value is less than zero, that means the function at that x value must be decreasing, meaning it is going down to the right, which is why it has a negative slope. So using that information, the first derivative test is a way of classifying a critical number as a local max, a local min, or neither. Now I should define for you what is a critical number. A critical number is a number, let's call it c, that is in the domain of f at x where f prime at c equals zero or is undefined. Let's focus in on if the critical number makes the first derivative zero. If it makes the first derivative zero, that means at that point the slope of the function is zero. So that could mean one of three things. It could be a local minimum point, it could be a local maximum point, or it could be neither. Let's zoom in on those three scenarios and look at how we can tell which of the three we have. Notice at all three of the points I've made, if I drew a tangent line they would all be perfectly horizontal, which is why the first derivative would be zero. So if all three scenarios result in a first derivative being zero, how do we know if a critical number is a min, a max, or neither? Well, we analyze the function before and after the critical point. If we look at this scenario where we have a local minimum point, notice that the function is decreasing before the critical point, but then increasing after the critical point. That would mean that the first derivative would be negative before the critical point and then positive after the critical point. In the scenario where there's a local maximum point, before the local maximum point the function's increasing, and then after the local max it's decreasing. So the first derivative would be positive before the critical point and then it would be negative after the critical point. If the critical point is not a local max or a min, the function is not going to switch between increasing and decreasing. For example, it could be increasing before the critical point and continue increasing after, so the first derivative will not change signs. It's positive before the critical point, and it's positive again after the critical point. So the first derivative test is just a test which is an organized way of testing which of those scenarios do you have. So let's do an example where it says to find all of the local extrema of f at x equals 2x cubed plus 3x squared minus 36x plus 5. For this polynomial function, I know at any local extrema, so at any local max or min point, the slope of the function is going to be zero. So I'm going to start by finding the critical numbers of this function by finding the derivative, setting it to zero, and solving. So I'll find the derivative f prime of x by differentiating each of the terms independently, which would give me 6x squared plus 6x minus 36. Now I want to know when is the slope of the function zero. So I'm going to set the derivative, which represents the slope of the function, equal to zero, and then solve the equation. I notice all three terms over here. I could common factor a 6 from all three terms. And now this quadratic is a trinomial with a leading coefficient of 1. So I just have to find numbers who have a product of negative 6 and a sum of 1. The numbers that satisfy that product and sum are positive 3 and negative 2. And now using the zero product rule, I know the product of these factors would be zero if either x plus three was zero or if x minus two is zero. And those factors would be zero if x was either negative three or two. So those numbers are my critical numbers. I'm going to test the value of the first derivative on each side of both of those critical numbers to see if the function is increasing or decreasing, and that will allow me to conclude. if I have a local max or a local min or neither at each of those x values. So to do that I'm actually going to set up a table. On the far left is always negative infinity, on the far right is infinity, and then between them is the critical numbers in ascending order, so negative three and two. What I'm doing is splitting those two critical numbers into one, two, three intervals, and then within each interval I'm going to find the sine of the first derivative and then draw a conclusion about what that tells me about the original function. So I now need to pick some test values. I'll start by finding a number between negative infinity and negative 3. I'll use negative 5. In my second interval between negative 3 and 2, I'll use 0. And from 2 to infinity, I'll use 5. I'll sub each of those test values into my first derivative, and just figure out what is the sign of the first derivative. When the first derivative is positive, negative... I know the original function has a positive slope, meaning it is increasing, it's going up to the right. And when the first derivative is negative, I know the original function has a negative slope, meaning it is decreasing, it is going down to the right. So what I see now from this table is that if I look at the critical number of negative three, the function was increasing before negative three, then decreasing after negative three, that corresponds to this scenario, where we have a local max. So I could say there is a local max at negative three, and then whatever the value of the function is at negative three. If I sub negative three into my f of x function, I would actually get 86. And then if I look at the critical number of two, before that critical number, the function was decreasing, then after it was increasing. So that corresponds to this scenario where there is a local min. So there's a local min at two, and then if I sub two into my f of x function, I get negative 39. So that's the first derivative test. Must know number five, the second derivative test. Before showing you the second derivative test, you have to know what the second derivative tells you about an original function. If the value of the second derivative at an x value is positive, so if it's greater than zero, the original function is what we call concave up. For a function to be concave up at a point, that would mean that at that point, if you were to draw a tangent line, the function would open above that tangent line. And notice concave up looks like a smiley face, so that's an easy way to remember that. And if the second derivative is negative at a given x value, we know that the original function is concave down at that x value. For a function to be concave down, it looks like this. Drawing a tangent line at any point on that shape, the function would open below that tangent line. So we say it's concave down. And also notice looks like a frowny face. So if we have a critical number, which is a number that makes the first derivative be zero, so let's say if we have f prime at c equals zero, the sine of f double prime at c can actually help us classify the point c f at c as a local max or a local min. Now let me draw the shape of a function that has a local max and a local min, so you can see how this would work. Notice at the local minimum point of this function, the slope of the function is zero. So the first derivative is zero. and the function is concave up, meaning the second derivative is positive. And at the local maximum point, the slope is again 0, so the first derivative is 0, but this time the function is concave down, meaning its second derivative would be negative. So the second derivative test is just a way of testing this. We find a number that makes the first derivative be 0, and then check and see, is the second derivative positive or negative? at that critical number, which will tell us if it's a local min or a local max. Now one thing that I should note for you is that if the first derivative is zero, and the second derivative is zero at an x value, that means that the second derivative test fails. And you have to use the first derivative test to try and classify it as a local max or a local min. So let's do an example using the second derivative test. Let's find the local extrema of f of x equals x cubed. minus 3x squared plus 2. For a polynomial function, any local extrema is going to have a slope of 0. So I'll start by finding the slope equation, which is the derivative f prime of x equals, if I differentiate all three terms, I get 3x squared minus 6x plus 0. And now I'll set this derivative, which is the slope of the function, equal to 0, and then solve for what value of x has that slope of 0. So I could solve this equation. I could common factor a 3x from both terms, giving me x minus 2. And then the product of those two factors would be 0 if either this factor is 0 or that factor is 0. And that would happen if x was equal to either 0 or 2. So I have two critical numbers. At those two x values, the slope of the function is 0. So it might be a max, it might be a min, it might be neither. I'll use the second derivative test to help me classify. I'll need to start by of course finding the second derivative by differentiating the first derivative. So the derivative of 3x squared minus 6x would be 6x minus 6. And now what I have to do, I have to check what is the value of this second derivative at each of these critical numbers. And that will tell me if the function is either concave up or concave down at each of those critical numbers. So I will evaluate f double prime at zero, and when I sub zero in for this x, I get negative six, and because this is negative, it tells me at the x value of zero, the original function is concave down. So since I know at an x value of zero, the slope is zero, and it's concave down, I know it must look like this where we have a local max point. So there's a local max at zero, and then whatever the value of f at 0 is. If I sub 0 into my f at x function, I get 2. Let's check the second critical number. Let's sub 2 into the second derivative. f double prime at 2 is equal to 6. Since this is positive, I know the original function is concave up when x is 2. Since it's concave up and it has a slope of 0, I know it must be this shape at that point, so it must be a local min. So there's a local min at 2 f at 2. When I sub 2 into the original function, I get negative 2. So 2 negative 2 is a local min. So that's how you use the second derivative test to find the local extrema of a function. Must know number six, curve sketching. Let's use everything we know about calculus so far to sketch a very accurate graph of the function f at x equals x cubed plus 6x squared plus 9x. Step one of the curve sketching process is to state any restrictions on the domain of the function and the equation of any asymptotes that it has. Because this is a polynomial function, I know it's continuous so there are no restrictions on the domain or any asymptotes. Let's move on to step number two. Step two is to find the x and y intercepts of the function. To find the x intercepts of a function, all you have to do is set the y coordinate equal to zero and then solve for what value of x makes the y coordinate be zero. To solve when a function is equal to zero, you want to get it into factored form. So I would common factor an x from all three terms. And then I noticed this quadratic in brackets. I could factor by finding numbers that multiply to nine and add to six. Those numbers are three and three. So it would factor to x plus three. times another factor of x plus 3, so I could just write it as x plus 3 squared. The product would be 0 if x was 0, or if x was negative 3. So my x-intercepts are at 0, 0, and negative 3, 0. To find the y-intercept of a function, all you have to do is set x equal to 0 and solve. And if I do that, I get 0. So the y-intercept is the point 0, 0, which I see I actually already had there as well. Let me make some room for step number 3. Step number three is to find the critical numbers of the function. Critical numbers are numbers that make the derivative of a function be zero or undefined, so I'll have to start by finding the derivative of the function, which would be 3x squared plus 12x plus 9. That's a continuous function so it can't be undefined, but it could be zero. Let me solve for when it's zero by common factoring a 3 from all of the terms, and then factoring this product by finding the numbers that multiply to 3 and add to 4. Those numbers are 3 and 1. So this would factor to x plus 3 times x plus 1. The product would be 0 if x was negative 3, or if x was negative 1. So my critical numbers are negative 3 and negative 1. I could find their corresponding y coordinates as well by taking those values and summing them into the original function for x. And I would get the critical points of negative 3, 0, and negative 1, negative 4. Let me shrink this and make room for the rest of the process. Step four is to find the possible points of inflection of the function. A point of inflection is a point on a function where it switches concavity. So this point right here is a point of inflection because before that point, the function's concave up and after it's concave down. So it switches concavity at that point. And at a point of inflection, we know the second derivative is going to be equal to zero or be undefined. So we'll find the second derivative. by differentiating the first derivative which was 3x squared plus 12x plus 9, so the second derivative would be 6x plus 12, and we'll want to solve for when is this equal to 0. If I solve this equation for x, I get negative 2. And if I sub x equals negative 2 into the original function, I actually get negative 2. So at the point negative 2, negative 2, the function might switch concavity. We'll have to test on either side of that point to see if there's a change in concavity. So let me make some room for us to do that test. In this last step, before we actually graph the function, we're going to find the intervals of when the function is increasing or decreasing and when the function is concave up or concave down. And we'll do that by picking test values on either side of any vertical asymptotes, critical numbers, or possible points of inflection we have for the function. So I'm going to make a chart where I'm going to test around all of those values. Alright, let me zoom in on this chart that I've made. Notice the intervals always start and end with negative and positive infinity, and then between there, we have both critical numbers and the possible point of inflection in ascending order. I'm going to pick a test value around each of those values. For each of those test values, I'm going to sub them into the first and second derivative and record the sign of the answer. When I sub negative 5 into the first derivative, I get a positive result. That means the function has a positive slope, which means it is increasing. So I can say the original function in that interval must be increasing. When I sub negative 5 into the second derivative, I get a negative result. That means that the original function must be concave down in that interval. So a function that is increasing and concave down looks like that shape. And then I'll do that same process for the remaining three test values. Just sub them into the first and second derivative and record the sign of the answer. So I'll do that quickly for you. And then remember the sign of the first derivative tells you whether the function is increasing or decreasing, and the sign of the second derivative tells you if it's concave up or down. So in this second interval between negative 3 and negative 2, the original function must be decreasing and concave down. Next interval, it must be decreasing and concave up, and then the last interval increasing and concave up. So now we should actually be able to state any local max or min points we have and any points of inflection. At negative 3, the function switches from increasing to decreasing, so there must be a local max. And at negative 1, the function switches from decreasing to increasing, so there's a local min. And at negative 2, there is a change in concavity, which means there's a point of inflection at negative 2. Using all of this information in the chart, we should be able to sketch a very accurate graph of this function. And we could actually just imagine if we pieced all four of those shapes together, we would get the exact graph of this function. So I'll zoom out. I'll make some room for me to do a graph. So there we have a rough sketch of this graph. Must know number seven, optimization. For this section, let me give you some context as to why we might be interested in finding the max and min values of a function. An open top box is to be made so that its base is twice as long as it is wide. The volume is 2,400 centimeters cubed. Find the dimensions that minimize the amount of cardboard used. Let me start by making a rough sketch of this box. Now, the question tells me that... The base is twice as long as it is wide, but I don't know how wide it is, so I'm going to assign the width the variable x, and the length has to be double that, so the length would be 2x. And I don't know the relationship between the height, length, and width, so I'll just call my height h for now. And because I'm interested in minimizing the amount of cardboard used, I want to minimize the surface area, so I'm going to need a formula for the surface area of this open top box. surface area, I just add up the area of all the sides. The front and the back are both x by h. So I would do 2 times x times h to get the areas of those two sides, plus let me look at the right side and the left side, which are both 2x by h. So 2 times 2x times h, plus the area of the bottom of this box. which is x by 2x. And because this is an open top box, I'm not considering the area of the top of it. If I simplify this surface area formula, I've got 2xh plus 4xh, that's 6xh, plus x times 2x is 2x squared. What I haven't done yet is to take into account that this box has to have a volume of 2,400 centimeters cubed. That's what we call the constraint for this question. So let's jot down that constraint. And remember the volume of a rectangular base prism is just length times width times height. So the volume would be equal to 2x times x times h, and the volume has to be 2400 centimeters cubed. So 2400 equals 2x squared h. And now that I have an equation that relates x and h to each other, I can rewrite this function to be written in terms of one variable. So I'm actually going to rewrite the h to be written in terms of x. And I can do that if I isolate this equation for h. And if I zoom in, that would give me h equals 2400 over 2x squared, which is 1200 over x squared. So what I can do with that, I can take what h is equal to and replace the h in the surface area formula with that value. So then my surface area formula is written only in terms of x. I can say SA at x equals 6x times 1200 over x squared plus 2x squared. Simplify this equation a little bit. That x will cancel with one of those. And I can do 6 times 1200, which is 7200. So this surface area formula is very specific to the open top box that we have here. And it'll keep the volume constant at 2400 centimeters cubed. What we want to do is find the minimum value of this function. I know at a minimum value, the slope is going to be 0. So what I can do is I can differentiate this function, set it to 0 and solve to find where a minimum point might be. Before I do that, I'm going to rewrite this. I'm going to take this power of x, move it to the top by changing the sign of its exponent, and then I'll differentiate this. And now I need to set it to 0 and solve to find any critical numbers of this function. I'll move this term over. I'll multiply the x squared over and divide the 4 over. And let me make a little bit of room. To isolate x, I can just raise both sides to the power of a third. And I figure out that x is about 12.16 and the units we're in are centimeters. Now if x is 12.16, I can use this equation that I made here, sub in this x value right into here, and then be able to solve for what h would be if x is 12.16. If I do that, I get an h value of 8.12 centimeters. Now to verify that this x value actually makes the surface area be a minimum, we should do a first or a second derivative test. I'll do that quickly. I'll do a second derivative test. So for the second derivative test, I would have to calculate the second derivative by differentiating the first derivative. And then I'll sub in the critical number into the second derivative, and I get about 12. The fact that that's a positive number means that the surface area function is concave up at that point, which means the critical number we have must be a minimum point. So now that we've done that second derivative test, we could write our final answer. The question asked what are the dimensions that minimize the surface area, so I should clearly state the dimensions. Well the width was just x, so the width is 12.16. The length is double x, so if I double x I get 24.32. And the height we calculated was 8.12. So there you go, that's how an optimization question works. Must know number eight, antiderivatives, which is exactly what it sounds like. This is going to be the opposite of differentiation. And there's a very specific notation for this. Let me show you the notation then explain to you what it means. The integral of f of x dx is equal to capital f of x plus c. So in this notation, we have a lot going on. This right here is the integral symbol. This function here f of x is what we call the integrand dx. is the variable of integration, and this integral is equal to capital F of x, which we call the antiderivative, plus any constant value, we call this value c, we call this our constant of integration. Now the most important thing to take away about this equation I just wrote here, is that this capital F of x is the antiderivative of lowercase f of x. That means that if I were to differentiate this, I would get this. So let's make a note of that. So in doing the indefinite integral of some function, what we're doing is we're looking for a function that if we were to differentiate it, we would get that integrand. And we always have to throw a plus c because the derivative of any constant is zero, so that c value could be anything. There are lots of different techniques for calculating antiderivatives, but let me show you a couple of them. The first example, let's just calculate the integral of 3x to the 4 dx. What that means is I want to find a function that if I differentiate it, I get 3x to the 4. For this integral, all we would need is the power rule of integration. The integral of any power function, so x to the n dx, will be equal to, you increase the exponent by 1, so x to the n plus 1, and then divide by n plus 1. And don't forget to add the constant of integration, c. So I'll use that rule for this question. Also, Anytime you have a constant factor in part of your integrand, you can move it in front of the integral symbol. And then using the power rule of integration, I know the integral of x to the 4. Well, I just have to raise the exponent by 1 and then divide by that new exponent. And then also don't forget to write that constant of integration c at the end. Another important method for calculating the indefinite integral of a function is u-substitution. Let's say we wanted to calculate the integral of sine of x squared plus 5 times x dx. If you have a composite function like this, it might be helpful if you rewrite the inside function, so x squared plus 5, if you rewrite it equal to u. And then we can try and write everything in terms of u. And you would be able to do that if we then found the derivative of u with respect to x. So I could say du over dx would be equal to Well if I differentiate this, I just get 2x. That would mean that du equals 2x dx. And what I notice now is that I've got an x dx here, but a 2x dx here. I can make this be x dx if I multiply both sides of that by a half. So now I'll be able to write this integral entirely in terms of u. I can replace x dx with a half du, and I can replace x squared plus five with u. This constant factor can move in front. And then calculating this integral is a lot easier. I need to find a function whose derivative is sine u. Well the derivative of cos is negative sine, so the derivative of negative cos will be positive sine. So this would equal a half times negative cosine of u, which I could write as negative a half cos u, but remember u is actually x squared plus 5. And there we go. There's the antiderivative, but don't forget, always have your plus c. You need the constant of integration at the end of any integral that you do. Must know number nine, definite integrals. Now that you know what antiderivatives are, let's see one of their main applications, which is finding an area under a curve. Let's say we're doing an example where we have to find the area of the region bounded by the graph of f at x equals negative x squared plus 5, the x-axis, and the vertical lines x equals 0 and x equals 2. So let me draw a graph of this so we can visualize what this is asking. So looking at this diagram, what the question is saying to find, it's saying find the area bounded by the region of our function, the line x equals 0, x equals 2, and the x-axis. So we're looking to find this whole area here, which is a difficult calculation because of that curved side. But what we can do, let me zoom in a little bit. We can break up this interval between 0 and 2. We can break it up into a number of subintervals. Let me break it into five sections. Within each of these five intervals that I've created, I'm going to draw a rectangle. And if I look at this first interval I've created, what I'm going to do to draw the rectangle is I'm going to use the right side of that interval. I'm going to go up and find where is the function at the right side of that interval and use that as the height of my rectangle. And I'll do that same process for the other four rectangles. Now notice those rectangles do not completely fill this area that I want, but it's a good estimation. Let me actually number these rectangles that I've had. Let me index them. This is my first rectangle, second, and so on. And then to calculate the area of any of these rectangles, what I would do is I would find the change in x between the left and the right side of the intervals. So I'll call this delta x, and I would find the height of the rectangle. I'll call this f at x sub i, and I'm calling it f at x sub i because the height is based on the right side of each of these intervals. And I could label the right side of the first interval x sub 1, the second rectangle x sub 2, and so on. And now I could write a general formula for finding the area underneath of a curve based on the idea that we're just adding up the area of rectangles. So this area that I'm trying to find here, I could say is equal to the sum from i equals one to five, right, I'm going to add up all five areas of the rectangles. And for the area of each rectangle, the area of a rectangle is just its base times its height. So it would be f at x sub i times delta x. The problem with this formula though is there is lots of missing space here. So that's not the actual area of what we're looking for. In order to fill in more of this space, we could break up this interval into more rectangles. And what would make it the most accurate is if we broke it up into an infinite number of rectangles. So we'll actually find the limit of the sum of the areas of rectangle one, all the way up to rectangle n, where n is approaching infinity of remember the area is f at x sub i times delta x that's just its base times its height and there's a notation that calculus uses to communicate this infinite sum of areas we could say that this area is equal to the definite integral from a to b we need to lower in an upper boundary which will eventually replace with zero and two once i apply it to this function of f at x dx so this and this are communicating the same thing. This sigma means sum, the same way this integral symbol, which kind of looks like an s, you can also think of it as a sum. We're finding the sum of the areas of a function times infinitely small values of x between a lower and an upper boundary. So we're just finding an infinite sum of rectangles under the curve. And based on the fundamental theorem of calculus, I know this is equal to the antiderivative at the upper boundary minus the antiderivative at the lower boundary. So let's apply this formula to the function that we have here and actually calculate that area we were looking for. So let me make some room and let's actually calculate the area between zero and two. So we'll find the definite integral between zero and two of our function, which is negative x squared plus five dx. So to evaluate this, I need to evaluate the antiderivative of that at 2 minus its value at 0. So the antiderivative of negative x squared plus 5 would be negative x cubed over 3 plus 5x. And my limits for this are 0 and 2. So I'll sub 2 in for these x's and then subtract what I get when I sub 0 in for those x's. And that would give me negative 2 cubed. over 3 plus 5 times 2 minus 0. So negative 8 thirds plus 10, which is 22 thirds, or about 7.33 units squared. Must know number 10, calculating the volume of a solid of revolution using the disk method. This is another application of a definite integral. In this example, we're going to find the volume of the solid obtained by rotating about the x axis, the region under the curve y equals root x from 0 to 2. So let me draw a diagram of this so you could visualize the problem. So I've got the function y equals root x, and what I want to do is rotate this around the x-axis to form a solid of revolution. So I'll rotate it around the x-axis. Now if I want the whole volume of this three-dimensional shape, what I could do is try and break this into a number of disks and then add up the volumes of those disks. Let me draw you a couple of those such disks. Now let me zoom in on what I drew here, and you can see that I drew two disks. inside of this solid of revolution, I drew one at that x value and that x value, you'll have to imagine that I was able to break this into an infinite number of disks and then I could find the volume of this whole shape by adding up the volumes of all of the disks that make up that shape. Now the disks that I drew actually by a very little bit overestimate the volume of this shape, but by breaking it up into an infinite number of disks, making that change in x, the kind of height of that cylinder as that approaches zero we're going to get an accurate estimation for the volume of that shape so i could communicate this a couple different ways i could say that the volume of that solid of revolution would be equal to the limit as the number of disks approaches infinity of the sum of disk one all the way up to disk n of the volume of those disks. And how do I find the volume of those disks? Well those disks are cylinders and I can find the volume of a cylinder by doing the area of the base of that cylinder, so the area at x sub i, and multiply that by the height of the cylinder, which is the change in the x value. So times delta x. And writing this infinite sum in calculus we can write that as a definite integral. I can say that the volume is equal to the integral from a to b of the area of the face of the disk times the height of the disk, where this dx implies that we have infinitely small changes in x, which is going to make these disks have infinitely small heights, which is going to break it up into an infinite number of disks. So to apply that to this question, well, first of all, my boundaries of integration are zero and two. and then to find the area of the face of each of those disks. Well it's a circle so I'm going to have to do pi r squared. Well what is the radius of each of these circles? Well notice the radius of that circle is just equal to the value of the function at that x value. So I could draw this radius here and notice that that's just equal to root x. So I'll zoom back out and I could replace the area of the face of the disk so the area of that circle as just pi times the radius squared. So it'd be pi times root x squared dx. And then the square root of x squared is just x. And then I can find the antiderivative of pi x just using the power rule of integration, it would be pi over two x squared. And I'm going to have to evaluate it at two minus its value at zero. So that'd be pi over two times two squared. minus pi over 2 times 0 squared, so minus 0. So this total volume would just be equal to 4 pi over 2, which is 2 pi. Let me zoom out so you can see everything that we did there, and if you followed along through that whole video, hopefully you have a good understanding of all the most important parts of calculus.

f of x equals x squared minus 2x plus 4, and then we'll use the derivative to evaluate f prime of 4. Now the derivative of a function is an equation that can calculate the slope or instantaneous rate of change of a function at a single point. I'll use this diagram that I've sketched of this f of x function to help explain this. Let's say for this function f of x that I was interested in finding the slope of it at a single point at some x value. Let's put the point right there. I could label the coordinates of that point, point x f at x.

The slope of the function at that single point could be visualized by me drawing a line that is tangent to this function at that point. The slope of this tangent line represents the slope or instantaneous rate of change of f at x at that point. But the challenging part of this is, how do we find the slope of a line where we only know one point that's on the line? Slope is usually calculated using two points and then finding their change in y and then dividing that by their change in their x values.

So what we could do is pick another point that's on this f at x function. I'll put the point right here. And let's say that that point is h units to the right of the original point. So its x coordinate would be x plus h and its y coordinate would be the value of the function at x plus h. And now it would be easy to calculate the slope of the line that connects those two points.

We can find the slope of that secant line just by doing the change in the y coordinates f at x plus h minus f at x, and then dividing that by the change in the x coordinates x plus h minus x. And those x's would cancel out and in the denominator we'd be left with just an h. Now the slope of that secant line does not match the slope of that tangent line. We could make it a better match or a better approximation by moving that point closer along the function to the original point that we want the slope at.

So we can make the h value, which stands for the change in x, we could make it smaller. So let me zoom in to show you how that would work. If I put the point right here and then found this secant line, notice that that secant line more closely matches the slope of the tangent line.

So the logical thing to do to get the best approximation for the slope of this tangent line is to move this point. infinitely close to the point we want the slope at. And we have a notation for that.

f prime of x. This apostrophe means a lot. It means it's the equation of the derivative, which means it's finding the slope of the tangent at a point.

It's the instantaneous rate of change of the function at a point. And to find that, what we want to do is find the limit of the secant slopes, so of that formula, as the change in x, which we call h, approaches zero. So this just communicates that as this point moves infinitely close to the point we want the slope at, the slope of the secant line approaches the slope of the tangent line.

And now let's use this formula to find the derivative of that specific f function. So f prime of x equals the limit as h goes to zero of f to evaluate a function at x plus h, I need to sub x plus h in for both of the x's into the function. And then Newton's quotient tells me I then have to subtract f at x.

So I need to subtract that function. And that all needs to be divided by h. Now this limit is in indeterminate form right now. We can't do direct substitution yet, but we can expand the numerator, collect our like terms, and get this out of indeterminate form. So I'll do x plus h all squared.

I'll distribute the negative 2 to both terms, and I'll subtract all. all three terms of f of x. And now let me zoom in so we can see what we can simplify here. x squared minus x squared, those cancel out.

Negative 2x plus 2x equals 0, and 4 minus 4 equals 0. Of all three terms that are left in the numerator, notice they all have an h. So on my next line, I'm going to common factor out that h. And now I know h is actually not going to be equal to 0 because we're finding the limit as h approaches 0. So I know that I can cancel out those h's, and I'll be left with a limit that is equivalent to this limit. And now this limit is out of indeterminate form, so now I can do direct substitution. I can sub 0 in for h, and then I have the equation of my derivative, which of course would simplify to just 2x minus 2. So this equation which we call the equation of the derivative, can find you the slope or instantaneous rate of change of the f at x function at any x value you want.

And in this question, it's said to evaluate f prime at 4. That means find the slope of the function at an x value of 4. f prime at 4, we just sub 4 in for x, 2 times 4 minus 2 is 8 minus 2, which is 6. So this means at an x value of 4, the slope. or instantaneous rate of change of the f of x function is 6. Must know number two, derivative rules. Not all derivatives have to be calculated from first principles using Newton's quotient.

There are four basic rules which will save you a lot of time. There's the power rule, the product rule, the quotient rule, and the chain rule. Let's start with the power rule. If you're trying to find the derivative, which I could denote as d over dx, of a power function, so x to the n, there's a rule that says you can write the exponent as the coefficient of the power and then reduce the exponent by 1. So the derivative of x to the n with respect to x would be n times x to the n minus 1. And let me do an example to show you how this rule works.

Let's find the derivative with respect to x of x to the power of 5. The power rule says take the exponent, write it as the coefficient, and then reduce the value of the exponent by 1. So this becomes 5 times x to the power of 4. Let's try another example. Let's find the derivative with respect to x of 3 times x to the 5. Well when we have a constant factor as part of our derivative, we're actually allowed to move that in front of our derivative operator and then just multiply that by the derivative of the power. And we know the derivative of x to the 5. We write the exponent as the coefficient, keep the base, and reduce the exponent by 1. And then these constant factors of 3 and 5 can be multiplied together to give us 15x to the power of 4. So you can probably see the shortcut that that exponent can just be multiplied by whatever coefficient is already there. 5 times 3 is 15. And then the exponent, 5, reduces by 1 to give you 4. Let's move on to the next rule I want to show you, the product rule.

Let's say you're finding the derivative with respect to x of a product of two functions of x. Let's call those functions f at x and g of x. Think of f at x as the first function of the product and g of x as the second function of the product.

The product rule says do the derivative of the first function, so f prime of x, multiply that by the second function, plus do the derivative of the second function, and then multiply that by the first function. Let's see how that works. Let's find the derivative with respect to x of x squared plus three times three x minus one. Notice we have a product of two functions of x, x squared plus three is our first function of x three x minus one is our second function of x.

If I follow the product rule, it tells me to do the derivative of the first factor. So the derivative of x squared plus three is two x. that gets multiplied by the second factor plus the derivative of the second factor, the derivative of 3x minus 1 is 3, and that gets multiplied by the first factor. So I'll expand this into standard form by distributing the 2x and the 3, and then collecting my like terms gives me 9x squared minus 2x plus 9. The next rule, the quotient rule, which is very related to the product rule, but this time what if I want to find the derivative with respect to x of a quotient of two functions of x? I have a top function and a bottom function.

The quotient rule says do the derivative of the top times the bottom function minus the derivative of the bottom function times the top function. And that all has to be divided by the bottom function squared. And then let me make some room here.

So let me shrink this rule and we'll do an example. Let's find the derivative of 3x divided by x squared plus 1. Because I have a function of x being divided by another function of x, I need to use the quotient rule, which tells me do the derivative of the top function, the derivative of 3x is 3. That gets multiplied by the bottom function, minus the derivative of the bottom function is 2x, multiplied by the top function. And then don't forget this all gets divided by this bottom function squared. And if I expand the numerator into standard form by distributing this 3, finding this product, and then collecting my like terms, I would get negative 3 x squared.

plus 3 divided by that x squared plus 1 squared. And lastly let's learn the chain rule, which is a rule we use when we have composite functions. So let's say we want to find the derivative with respect to x of f of g of x. So it's a function of a function. When you have that, you differentiate the outside function and leave the inside function exactly as it is. but then that has to be multiplied by the derivative of the inside function.

So we'd multiply that by g prime of x. Now this chain rule can be applied to lots of different types of functions, but because for these examples we've been focusing on power functions, let me show you how this chain rule specifically applies to power functions. So let's say we're finding the derivative of not just x to the power of n, but f of x to the power of n. Using the power rule, you know we have to write that exponent as the coefficient of the power. We'll leave the base of the power exactly as it is, and then we'll reduce the exponent by one.

So we'd have n times f of x to the n minus one. But because the base of the power is a function of x, and it's not just x, we then have to follow a chain rule which says multiply by the derivative of the inside function. So I need to multiply by the derivative of the base of the power. So this would be multiplied by f prime of x. So let's make some room and do an example to show you exactly how this would work.

Let's find the derivative with respect to x of 3x minus 2 to the power of 4. This is a power function where the base of the power is a function of x. So following the chain rule, I know I take the exponent, write it as the coefficient of the power. I leave the base of the power exactly as it is.

I reduce the exponent by 1. But then I have to remember to multiply this by the derivative. of the inside function. The derivative of 3x minus 2 is 3. And then I could simplify this by multiplying those constant factors together. 4 times 3 is 12. So I have 12 times 3x minus 2 cubed. So let me zoom out here so you can see all four rules that we learned.

Must know number three, derivatives of other types of functions. There are three other important types of functions that you need to know how to find the derivative of. Exponential functions, logarithmic functions, and of course trig functions. Let's start with exponential functions. When finding the derivative of an exponential function, so the derivative with respect to x of b to the power of x would be equal to, you keep the power, so b to the power of x stays the exact same, but that gets multiplied by the natural logarithm of the base of the power.

So for example, when finding the derivative of five to the power of x, that's just equal to, you keep the power exactly as it is, five to the power of x, and multiply that by the natural logarithm of the base of the power, which is five. And then what if the exponent isn't just x? What if it's a function of x? So let's find the derivative with respect to x of b to the power of f of x.

You would need the chain rule. keep the power exactly as it is, still multiply by the natural logarithm of the base, but that would then have to be multiplied by the derivative of the function that's in the exponents. We'd multiply this by f prime of x.

Let me make some room for an example, and let's find the derivative with respect to x of 5 to the power of 3x plus 1. Keep the power times the natural logarithm of the base, but then also multiply... by the derivative of the function that's in the exponent. The derivative of 3x plus 1 is 3. And then we'll just simplify that expression by moving those constant factors to the front.

Let's move on to logarithmic functions. The derivative of log base b of x is equal to 1 divided by the argument of the logarithm, so x, times the natural logarithm of the base of the logarithm, which is b. An example where we could practice that would be finding the derivative of log base 5 of x.

That would be equal to 1 divided by x times the natural logarithm of 5. And then what if the argument wasn't just x? What if it was a function of x? To find the derivative of log base b of f at x, you would need to use the chain rule, which would say follow the same rule 1 divided by the argument which is f times the natural logarithm of the base b, but then you have to multiply all of that by the derivative of the inside function. So multiply by the derivative of f.

An example of that would be finding the derivative of log base 5 of 2x cubed. Following the derivative rule, I would do 1 divided by, keep the argument the same, 2x cubed, times the natural logarithm of the base of the log. But then that has to get multiplied by the derivative of the inside function.

The derivative of 2x cubed is 6x squared. And then I'll make a little bit of room so I can simplify this. Notice I have a 6 divided by 2, that's 3, and both of those factors of x could cancel with two of those factors of x, leaving me with 3 divided by x times the natural logarithm of 5. And lastly let's move on to trig functions. You need to know how to differentiate the three primary trig functions. The derivative of sin x is equal to cos x, the derivative of cos x is negative sin x, and the derivative of tan x is equal to secant squared of x.

And then an example to practice doing derivative rules that involve trig functions, let's find the derivative of 2 times sine of 3x. This constant factor can go in front of the derivative operator, so this would equal 2 times the derivative of sine of 3x. And then when differentiating a sine function, I know the derivative sine of anything is equal to cosine of that same argument.

But because the argument of the sine function was not just x, it was a function of x, this is a composite function which means I need chain rule. So I then have to multiply this by the derivative of 3x. The derivative of 3x is 3. And the last thing to do, those constant factors of 2 and 3 can be multiplied together to give 6. So it equals 6 times cosine of 3x. I'll zoom out so you can see all of those rules.

Must know number four, the first derivative test. Before I show you the first derivative test, you're going to have to remember what the first derivative tells you about a function. If the value of the first derivative at an x value is greater than zero, so if it's positive, that means that the original function at that x value is negative.

increasing. That's because if we have a positive first derivative, that means the slope is positive, which means the function must be going up to the right. If the value of the first derivative at an x value is less than zero, that means the function at that x value must be decreasing, meaning it is going down to the right, which is why it has a negative slope. So using that information, the first derivative test is a way of classifying a critical number as a local max, a local min, or neither. Now I should define for you what is a critical number.

A critical number is a number, let's call it c, that is in the domain of f at x where f prime at c equals zero or is undefined. Let's focus in on if the critical number makes the first derivative zero. If it makes the first derivative zero, that means at that point the slope of the function is zero.

So that could mean one of three things. It could be a local minimum point, it could be a local maximum point, or it could be neither. Let's zoom in on those three scenarios and look at how we can tell which of the three we have. Notice at all three of the points I've made, if I drew a tangent line they would all be perfectly horizontal, which is why the first derivative would be zero. So if all three scenarios result in a first derivative being zero, how do we know if a critical number is a min, a max, or neither?

Well, we analyze the function before and after the critical point. If we look at this scenario where we have a local minimum point, notice that the function is decreasing before the critical point, but then increasing after the critical point. That would mean that the first derivative would be negative before the critical point and then positive after the critical point.

In the scenario where there's a local maximum point, before the local maximum point the function's increasing, and then after the local max it's decreasing. So the first derivative would be positive before the critical point and then it would be negative after the critical point. If the critical point is not a local max or a min, the function is not going to switch between increasing and decreasing.

For example, it could be increasing before the critical point and continue increasing after, so the first derivative will not change signs. It's positive before the critical point, and it's positive again after the critical point. So the first derivative test is just a test which is an organized way of testing which of those scenarios do you have. So let's do an example where it says to find all of the local extrema of f at x equals 2x cubed plus 3x squared minus 36x plus 5. For this polynomial function, I know at any local extrema, so at any local max or min point, the slope of the function is going to be zero. So I'm going to start by finding the critical numbers of this function by finding the derivative, setting it to zero, and solving.

So I'll find the derivative f prime of x by differentiating each of the terms independently, which would give me 6x squared plus 6x minus 36. Now I want to know when is the slope of the function zero. So I'm going to set the derivative, which represents the slope of the function, equal to zero, and then solve the equation. I notice all three terms over here.

I could common factor a 6 from all three terms. And now this quadratic is a trinomial with a leading coefficient of 1. So I just have to find numbers who have a product of negative 6 and a sum of 1. The numbers that satisfy that product and sum are positive 3 and negative 2. And now using the zero product rule, I know the product of these factors would be zero if either x plus three was zero or if x minus two is zero. And those factors would be zero if x was either negative three or two.

So those numbers are my critical numbers. I'm going to test the value of the first derivative on each side of both of those critical numbers to see if the function is increasing or decreasing, and that will allow me to conclude. if I have a local max or a local min or neither at each of those x values.

So to do that I'm actually going to set up a table. On the far left is always negative infinity, on the far right is infinity, and then between them is the critical numbers in ascending order, so negative three and two. What I'm doing is splitting those two critical numbers into one, two, three intervals, and then within each interval I'm going to find the sine of the first derivative and then draw a conclusion about what that tells me about the original function.

So I now need to pick some test values. I'll start by finding a number between negative infinity and negative 3. I'll use negative 5. In my second interval between negative 3 and 2, I'll use 0. And from 2 to infinity, I'll use 5. I'll sub each of those test values into my first derivative, and just figure out what is the sign of the first derivative. When the first derivative is positive, negative... I know the original function has a positive slope, meaning it is increasing, it's going up to the right.

And when the first derivative is negative, I know the original function has a negative slope, meaning it is decreasing, it is going down to the right. So what I see now from this table is that if I look at the critical number of negative three, the function was increasing before negative three, then decreasing after negative three, that corresponds to this scenario, where we have a local max. So I could say there is a local max at negative three, and then whatever the value of the function is at negative three. If I sub negative three into my f of x function, I would actually get 86. And then if I look at the critical number of two, before that critical number, the function was decreasing, then after it was increasing. So that corresponds to this scenario where there is a local min.

So there's a local min at two, and then if I sub two into my f of x function, I get negative 39. So that's the first derivative test. Must know number five, the second derivative test. Before showing you the second derivative test, you have to know what the second derivative tells you about an original function.

If the value of the second derivative at an x value is positive, so if it's greater than zero, the original function is what we call concave up. For a function to be concave up at a point, that would mean that at that point, if you were to draw a tangent line, the function would open above that tangent line. And notice concave up looks like a smiley face, so that's an easy way to remember that. And if the second derivative is negative at a given x value, we know that the original function is concave down at that x value. For a function to be concave down, it looks like this.

Drawing a tangent line at any point on that shape, the function would open below that tangent line. So we say it's concave down. And also notice looks like a frowny face. So if we have a critical number, which is a number that makes the first derivative be zero, so let's say if we have f prime at c equals zero, the sine of f double prime at c can actually help us classify the point c f at c as a local max or a local min. Now let me draw the shape of a function that has a local max and a local min, so you can see how this would work.

Notice at the local minimum point of this function, the slope of the function is zero. So the first derivative is zero. and the function is concave up, meaning the second derivative is positive.

And at the local maximum point, the slope is again 0, so the first derivative is 0, but this time the function is concave down, meaning its second derivative would be negative. So the second derivative test is just a way of testing this. We find a number that makes the first derivative be 0, and then check and see, is the second derivative positive or negative? at that critical number, which will tell us if it's a local min or a local max.

Now one thing that I should note for you is that if the first derivative is zero, and the second derivative is zero at an x value, that means that the second derivative test fails. And you have to use the first derivative test to try and classify it as a local max or a local min. So let's do an example using the second derivative test.

Let's find the local extrema of f of x equals x cubed. minus 3x squared plus 2. For a polynomial function, any local extrema is going to have a slope of 0. So I'll start by finding the slope equation, which is the derivative f prime of x equals, if I differentiate all three terms, I get 3x squared minus 6x plus 0. And now I'll set this derivative, which is the slope of the function, equal to 0, and then solve for what value of x has that slope of 0. So I could solve this equation. I could common factor a 3x from both terms, giving me x minus 2. And then the product of those two factors would be 0 if either this factor is 0 or that factor is 0. And that would happen if x was equal to either 0 or 2. So I have two critical numbers. At those two x values, the slope of the function is 0. So it might be a max, it might be a min, it might be neither. I'll use the second derivative test to help me classify.

I'll need to start by of course finding the second derivative by differentiating the first derivative. So the derivative of 3x squared minus 6x would be 6x minus 6. And now what I have to do, I have to check what is the value of this second derivative at each of these critical numbers. And that will tell me if the function is either concave up or concave down at each of those critical numbers. So I will evaluate f double prime at zero, and when I sub zero in for this x, I get negative six, and because this is negative, it tells me at the x value of zero, the original function is concave down. So since I know at an x value of zero, the slope is zero, and it's concave down, I know it must look like this where we have a local max point.

So there's a local max at zero, and then whatever the value of f at 0 is. If I sub 0 into my f at x function, I get 2. Let's check the second critical number. Let's sub 2 into the second derivative.

f double prime at 2 is equal to 6. Since this is positive, I know the original function is concave up when x is 2. Since it's concave up and it has a slope of 0, I know it must be this shape at that point, so it must be a local min. So there's a local min at 2 f at 2. When I sub 2 into the original function, I get negative 2. So 2 negative 2 is a local min. So that's how you use the second derivative test to find the local extrema of a function. Must know number six, curve sketching. Let's use everything we know about calculus so far to sketch a very accurate graph of the function f at x equals x cubed plus 6x squared plus 9x.

Step one of the curve sketching process is to state any restrictions on the domain of the function and the equation of any asymptotes that it has. Because this is a polynomial function, I know it's continuous so there are no restrictions on the domain or any asymptotes. Let's move on to step number two.

Step two is to find the x and y intercepts of the function. To find the x intercepts of a function, all you have to do is set the y coordinate equal to zero and then solve for what value of x makes the y coordinate be zero. To solve when a function is equal to zero, you want to get it into factored form.

So I would common factor an x from all three terms. And then I noticed this quadratic in brackets. I could factor by finding numbers that multiply to nine and add to six. Those numbers are three and three. So it would factor to x plus three.

times another factor of x plus 3, so I could just write it as x plus 3 squared. The product would be 0 if x was 0, or if x was negative 3. So my x-intercepts are at 0, 0, and negative 3, 0. To find the y-intercept of a function, all you have to do is set x equal to 0 and solve. And if I do that, I get 0. So the y-intercept is the point 0, 0, which I see I actually already had there as well. Let me make some room for step number 3. Step number three is to find the critical numbers of the function. Critical numbers are numbers that make the derivative of a function be zero or undefined, so I'll have to start by finding the derivative of the function, which would be 3x squared plus 12x plus 9. That's a continuous function so it can't be undefined, but it could be zero.

Let me solve for when it's zero by common factoring a 3 from all of the terms, and then factoring this product by finding the numbers that multiply to 3 and add to 4. Those numbers are 3 and 1. So this would factor to x plus 3 times x plus 1. The product would be 0 if x was negative 3, or if x was negative 1. So my critical numbers are negative 3 and negative 1. I could find their corresponding y coordinates as well by taking those values and summing them into the original function for x. And I would get the critical points of negative 3, 0, and negative 1, negative 4. Let me shrink this and make room for the rest of the process. Step four is to find the possible points of inflection of the function.

A point of inflection is a point on a function where it switches concavity. So this point right here is a point of inflection because before that point, the function's concave up and after it's concave down. So it switches concavity at that point.

And at a point of inflection, we know the second derivative is going to be equal to zero or be undefined. So we'll find the second derivative. by differentiating the first derivative which was 3x squared plus 12x plus 9, so the second derivative would be 6x plus 12, and we'll want to solve for when is this equal to 0. If I solve this equation for x, I get negative 2. And if I sub x equals negative 2 into the original function, I actually get negative 2. So at the point negative 2, negative 2, the function might switch concavity. We'll have to test on either side of that point to see if there's a change in concavity. So let me make some room for us to do that test.

In this last step, before we actually graph the function, we're going to find the intervals of when the function is increasing or decreasing and when the function is concave up or concave down. And we'll do that by picking test values on either side of any vertical asymptotes, critical numbers, or possible points of inflection we have for the function. So I'm going to make a chart where I'm going to test around all of those values.

Alright, let me zoom in on this chart that I've made. Notice the intervals always start and end with negative and positive infinity, and then between there, we have both critical numbers and the possible point of inflection in ascending order. I'm going to pick a test value around each of those values. For each of those test values, I'm going to sub them into the first and second derivative and record the sign of the answer.

When I sub negative 5 into the first derivative, I get a positive result. That means the function has a positive slope, which means it is increasing. So I can say the original function in that interval must be increasing.

When I sub negative 5 into the second derivative, I get a negative result. That means that the original function must be concave down in that interval. So a function that is increasing and concave down looks like that shape.

And then I'll do that same process for the remaining three test values. Just sub them into the first and second derivative and record the sign of the answer. So I'll do that quickly for you. And then remember the sign of the first derivative tells you whether the function is increasing or decreasing, and the sign of the second derivative tells you if it's concave up or down.

So in this second interval between negative 3 and negative 2, the original function must be decreasing and concave down. Next interval, it must be decreasing and concave up, and then the last interval increasing and concave up. So now we should actually be able to state any local max or min points we have and any points of inflection.

At negative 3, the function switches from increasing to decreasing, so there must be a local max. And at negative 1, the function switches from decreasing to increasing, so there's a local min. And at negative 2, there is a change in concavity, which means there's a point of inflection at negative 2. Using all of this information in the chart, we should be able to sketch a very accurate graph of this function.

And we could actually just imagine if we pieced all four of those shapes together, we would get the exact graph of this function. So I'll zoom out. I'll make some room for me to do a graph.

So there we have a rough sketch of this graph. Must know number seven, optimization. For this section, let me give you some context as to why we might be interested in finding the max and min values of a function.

An open top box is to be made so that its base is twice as long as it is wide. The volume is 2,400 centimeters cubed. Find the dimensions that minimize the amount of cardboard used. Let me start by making a rough sketch of this box.

Now, the question tells me that... The base is twice as long as it is wide, but I don't know how wide it is, so I'm going to assign the width the variable x, and the length has to be double that, so the length would be 2x. And I don't know the relationship between the height, length, and width, so I'll just call my height h for now. And because I'm interested in minimizing the amount of cardboard used, I want to minimize the surface area, so I'm going to need a formula for the surface area of this open top box.

surface area, I just add up the area of all the sides. The front and the back are both x by h. So I would do 2 times x times h to get the areas of those two sides, plus let me look at the right side and the left side, which are both 2x by h. So 2 times 2x times h, plus the area of the bottom of this box. which is x by 2x.

And because this is an open top box, I'm not considering the area of the top of it. If I simplify this surface area formula, I've got 2xh plus 4xh, that's 6xh, plus x times 2x is 2x squared. What I haven't done yet is to take into account that this box has to have a volume of 2,400 centimeters cubed. That's what we call the constraint for this question.

So let's jot down that constraint. And remember the volume of a rectangular base prism is just length times width times height. So the volume would be equal to 2x times x times h, and the volume has to be 2400 centimeters cubed.

So 2400 equals 2x squared h. And now that I have an equation that relates x and h to each other, I can rewrite this function to be written in terms of one variable. So I'm actually going to rewrite the h to be written in terms of x. And I can do that if I isolate this equation for h. And if I zoom in, that would give me h equals 2400 over 2x squared, which is 1200 over x squared.

So what I can do with that, I can take what h is equal to and replace the h in the surface area formula with that value. So then my surface area formula is written only in terms of x. I can say SA at x equals 6x times 1200 over x squared plus 2x squared.

Simplify this equation a little bit. That x will cancel with one of those. And I can do 6 times 1200, which is 7200. So this surface area formula is very specific to the open top box that we have here. And it'll keep the volume constant at 2400 centimeters cubed. What we want to do is find the minimum value of this function.

I know at a minimum value, the slope is going to be 0. So what I can do is I can differentiate this function, set it to 0 and solve to find where a minimum point might be. Before I do that, I'm going to rewrite this. I'm going to take this power of x, move it to the top by changing the sign of its exponent, and then I'll differentiate this.

And now I need to set it to 0 and solve to find any critical numbers of this function. I'll move this term over. I'll multiply the x squared over and divide the 4 over.

And let me make a little bit of room. To isolate x, I can just raise both sides to the power of a third. And I figure out that x is about 12.16 and the units we're in are centimeters. Now if x is 12.16, I can use this equation that I made here, sub in this x value right into here, and then be able to solve for what h would be if x is 12.16. If I do that, I get an h value of 8.12 centimeters.

Now to verify that this x value actually makes the surface area be a minimum, we should do a first or a second derivative test. I'll do that quickly. I'll do a second derivative test. So for the second derivative test, I would have to calculate the second derivative by differentiating the first derivative. And then I'll sub in the critical number into the second derivative, and I get about 12. The fact that that's a positive number means that the surface area function is concave up at that point, which means the critical number we have must be a minimum point.

So now that we've done that second derivative test, we could write our final answer. The question asked what are the dimensions that minimize the surface area, so I should clearly state the dimensions. Well the width was just x, so the width is 12.16. The length is double x, so if I double x I get 24.32.

And the height we calculated was 8.12. So there you go, that's how an optimization question works. Must know number eight, antiderivatives, which is exactly what it sounds like.

This is going to be the opposite of differentiation. And there's a very specific notation for this. Let me show you the notation then explain to you what it means.

The integral of f of x dx is equal to capital f of x plus c. So in this notation, we have a lot going on. This right here is the integral symbol.

This function here f of x is what we call the integrand dx. is the variable of integration, and this integral is equal to capital F of x, which we call the antiderivative, plus any constant value, we call this value c, we call this our constant of integration. Now the most important thing to take away about this equation I just wrote here, is that this capital F of x is the antiderivative of lowercase f of x.

That means that if I were to differentiate this, I would get this. So let's make a note of that. So in doing the indefinite integral of some function, what we're doing is we're looking for a function that if we were to differentiate it, we would get that integrand. And we always have to throw a plus c because the derivative of any constant is zero, so that c value could be anything.

There are lots of different techniques for calculating antiderivatives, but let me show you a couple of them. The first example, let's just calculate the integral of 3x to the 4 dx. What that means is I want to find a function that if I differentiate it, I get 3x to the 4. For this integral, all we would need is the power rule of integration. The integral of any power function, so x to the n dx, will be equal to, you increase the exponent by 1, so x to the n plus 1, and then divide by n plus 1. And don't forget to add the constant of integration, c. So I'll use that rule for this question.

Also, Anytime you have a constant factor in part of your integrand, you can move it in front of the integral symbol. And then using the power rule of integration, I know the integral of x to the 4. Well, I just have to raise the exponent by 1 and then divide by that new exponent. And then also don't forget to write that constant of integration c at the end.

Another important method for calculating the indefinite integral of a function is u-substitution. Let's say we wanted to calculate the integral of sine of x squared plus 5 times x dx. If you have a composite function like this, it might be helpful if you rewrite the inside function, so x squared plus 5, if you rewrite it equal to u. And then we can try and write everything in terms of u. And you would be able to do that if we then found the derivative of u with respect to x.

So I could say du over dx would be equal to Well if I differentiate this, I just get 2x. That would mean that du equals 2x dx. And what I notice now is that I've got an x dx here, but a 2x dx here. I can make this be x dx if I multiply both sides of that by a half. So now I'll be able to write this integral entirely in terms of u.

I can replace x dx with a half du, and I can replace x squared plus five with u. This constant factor can move in front. And then calculating this integral is a lot easier. I need to find a function whose derivative is sine u.

Well the derivative of cos is negative sine, so the derivative of negative cos will be positive sine. So this would equal a half times negative cosine of u, which I could write as negative a half cos u, but remember u is actually x squared plus 5. And there we go. There's the antiderivative, but don't forget, always have your plus c. You need the constant of integration at the end of any integral that you do. Must know number nine, definite integrals.

Now that you know what antiderivatives are, let's see one of their main applications, which is finding an area under a curve. Let's say we're doing an example where we have to find the area of the region bounded by the graph of f at x equals negative x squared plus 5, the x-axis, and the vertical lines x equals 0 and x equals 2. So let me draw a graph of this so we can visualize what this is asking. So looking at this diagram, what the question is saying to find, it's saying find the area bounded by the region of our function, the line x equals 0, x equals 2, and the x-axis.

So we're looking to find this whole area here, which is a difficult calculation because of that curved side. But what we can do, let me zoom in a little bit. We can break up this interval between 0 and 2. We can break it up into a number of subintervals. Let me break it into five sections. Within each of these five intervals that I've created, I'm going to draw a rectangle.

And if I look at this first interval I've created, what I'm going to do to draw the rectangle is I'm going to use the right side of that interval. I'm going to go up and find where is the function at the right side of that interval and use that as the height of my rectangle. And I'll do that same process for the other four rectangles.

Now notice those rectangles do not completely fill this area that I want, but it's a good estimation. Let me actually number these rectangles that I've had. Let me index them. This is my first rectangle, second, and so on.

And then to calculate the area of any of these rectangles, what I would do is I would find the change in x between the left and the right side of the intervals. So I'll call this delta x, and I would find the height of the rectangle. I'll call this f at x sub i, and I'm calling it f at x sub i because the height is based on the right side of each of these intervals.

And I could label the right side of the first interval x sub 1, the second rectangle x sub 2, and so on. And now I could write a general formula for finding the area underneath of a curve based on the idea that we're just adding up the area of rectangles. So this area that I'm trying to find here, I could say is equal to the sum from i equals one to five, right, I'm going to add up all five areas of the rectangles. And for the area of each rectangle, the area of a rectangle is just its base times its height.

So it would be f at x sub i times delta x. The problem with this formula though is there is lots of missing space here. So that's not the actual area of what we're looking for. In order to fill in more of this space, we could break up this interval into more rectangles. And what would make it the most accurate is if we broke it up into an infinite number of rectangles.

So we'll actually find the limit of the sum of the areas of rectangle one, all the way up to rectangle n, where n is approaching infinity of remember the area is f at x sub i times delta x that's just its base times its height and there's a notation that calculus uses to communicate this infinite sum of areas we could say that this area is equal to the definite integral from a to b we need to lower in an upper boundary which will eventually replace with zero and two once i apply it to this function of f at x dx so this and this are communicating the same thing. This sigma means sum, the same way this integral symbol, which kind of looks like an s, you can also think of it as a sum. We're finding the sum of the areas of a function times infinitely small values of x between a lower and an upper boundary.

So we're just finding an infinite sum of rectangles under the curve. And based on the fundamental theorem of calculus, I know this is equal to the antiderivative at the upper boundary minus the antiderivative at the lower boundary. So let's apply this formula to the function that we have here and actually calculate that area we were looking for.

So let me make some room and let's actually calculate the area between zero and two. So we'll find the definite integral between zero and two of our function, which is negative x squared plus five dx. So to evaluate this, I need to evaluate the antiderivative of that at 2 minus its value at 0. So the antiderivative of negative x squared plus 5 would be negative x cubed over 3 plus 5x.

And my limits for this are 0 and 2. So I'll sub 2 in for these x's and then subtract what I get when I sub 0 in for those x's. And that would give me negative 2 cubed. over 3 plus 5 times 2 minus 0. So negative 8 thirds plus 10, which is 22 thirds, or about 7.33 units squared.

Must know number 10, calculating the volume of a solid of revolution using the disk method. This is another application of a definite integral. In this example, we're going to find the volume of the solid obtained by rotating about the x axis, the region under the curve y equals root x from 0 to 2. So let me draw a diagram of this so you could visualize the problem. So I've got the function y equals root x, and what I want to do is rotate this around the x-axis to form a solid of revolution. So I'll rotate it around the x-axis.

Now if I want the whole volume of this three-dimensional shape, what I could do is try and break this into a number of disks and then add up the volumes of those disks. Let me draw you a couple of those such disks. Now let me zoom in on what I drew here, and you can see that I drew two disks. inside of this solid of revolution, I drew one at that x value and that x value, you'll have to imagine that I was able to break this into an infinite number of disks and then I could find the volume of this whole shape by adding up the volumes of all of the disks that make up that shape. Now the disks that I drew actually by a very little bit overestimate the volume of this shape, but by breaking it up into an infinite number of disks, making that change in x, the kind of height of that cylinder as that approaches zero we're going to get an accurate estimation for the volume of that shape so i could communicate this a couple different ways i could say that the volume of that solid of revolution would be equal to the limit as the number of disks approaches infinity of the sum of disk one all the way up to disk n of the volume of those disks.

And how do I find the volume of those disks? Well those disks are cylinders and I can find the volume of a cylinder by doing the area of the base of that cylinder, so the area at x sub i, and multiply that by the height of the cylinder, which is the change in the x value. So times delta x.

And writing this infinite sum in calculus we can write that as a definite integral. I can say that the volume is equal to the integral from a to b of the area of the face of the disk times the height of the disk, where this dx implies that we have infinitely small changes in x, which is going to make these disks have infinitely small heights, which is going to break it up into an infinite number of disks. So to apply that to this question, well, first of all, my boundaries of integration are zero and two. and then to find the area of the face of each of those disks.

Well it's a circle so I'm going to have to do pi r squared. Well what is the radius of each of these circles? Well notice the radius of that circle is just equal to the value of the function at that x value. So I could draw this radius here and notice that that's just equal to root x.

So I'll zoom back out and I could replace the area of the face of the disk so the area of that circle as just pi times the radius squared. So it'd be pi times root x squared dx. And then the square root of x squared is just x. And then I can find the antiderivative of pi x just using the power rule of integration, it would be pi over two x squared. And I'm going to have to evaluate it at two minus its value at zero.

So that'd be pi over two times two squared. minus pi over 2 times 0 squared, so minus 0. So this total volume would just be equal to 4 pi over 2, which is 2 pi. Let me zoom out so you can see everything that we did there, and if you followed along through that whole video, hopefully you have a good understanding of all the most important parts of calculus.

Transcript for:Fundamentals of Calculus Concepts

Transcript for:
Fundamentals of Calculus Concepts