We're about to review every single topic of AP Calc BC. This is the only video you're going to need to study for your exam. We're going to go through limits and continuity, the definition of differentiation, composite, implicit, and inverse differentiation, contextual applications of differentiation, analytical applications of differentiation, integration and accumulation of change, differential equations, applications of integration, parametric equations, polar coordinates, and vector value functions, and infinite sequences and series. Before calculus, we measure change using average rates, like average speed over a time interval. Calculus introduces the concept of instantaneous rates of change, how fast something is changing at an exact moment. This is where limits come in. Limits allow us to define the slope of a curve at a point. The foundation of derivatives. Limits follow predictable rules. You can break limits across addition, subtraction, multiplication, and division provided the individual limits exist. This allows for simplifying complex expressions and helps evaluate limits systematically. To find limits using graphs, observe what y-value the function approaches as x approaches a given point from both sides. If both sides match, the limit exists and equals that value. If they don't match or oscillate or approach infinity, the limit does not exist. Here we see an example of a function with the limit not existing at x= 0 because y = 1 /x is unbounded. On the right, we see a function that oscillates infinitely as x approaches zero, meaning the limit also does not exist. Using a table of values, look at the x values approaching a specific point from the left and the right. If the corresponding y-v values get closer to the same number from both sides, that number is the limit. This is useful when evaluating limits numerically. The squeeze theorem is used when a function is bounded between two others. If both bounding functions approach the same limit at a point, the squeezed function must also approach that limit. There are three types of discontinuities. Removable where there's a hole in the graph. Jump where the function leaps to another value. And infinite which occur near vertical asmmptotes. Identifying these is important for discussing continuity. For a function to be continuous at a point, three conditions must be met. The function must be defined at that point. The limit exists at that point. and the value of the function equals the limit. Continuity over an interval means the function is continuous at every point in that interval. As x approaches infinity or negative infinity, the behavior of a function is analyzed using limits at infinity. This helps determine horizontal asmmptotes and understand long-term trends of a function. The IVT states that if a function is continuous on the closed interval from a to b and a number lies between f of a and f of b, then there must be at least one value c in the open interval a to b where f of c equals that number. The derivative is defined by either of the limit definitions shown to the right. It has a few different notations. Frime of x, yprime, and dydx. Also, the derivative of a function is the slope of the tangent line at that point. When exact derivatives aren't available, you can estimate the slope of the tangent line at a point using tables or graphs. Just find the slope using two points near the x value. Differentiability implies continuity but continuity does not imply differentiability. A function is differentiable at a point if it is both continuous and smooth there. Sharp corners, cusps or discontinuities mean the derivative does not exist at that point. Basic rules make taking derivatives easier. The derivative of a constant is zero. You can add or subtract derivatives and multiply by a constant. The power rule applies to polomials. The product rule applies to the product of non-constant functions. The quotient rule applies to the quotient of non-constant functions. And the chain rule applies to functions inside of another function. Know the derivatives of common trig and exponential functions. The derivative of s is cosine. The derivative of cosine is negative sign. The derivative of e to x is itself. And the derivative of the natural log of x is 1 /x. You also have to memorize the derivatives of the other trig functions. The derivative of tangent is secant^ squ. The derivative of negative cosecant squ. The derivative of seccant is secant * tangent. And the derivative of coseant is cosecant. The chain rule is used when one function is nested inside another. This allows us to take derivatives of complex composite functions. Implicit differentiation is necessary when y cannot be easily isolated. Differentiate both sides of the equation with respect to x treating y as a function of x and then solve for dydx. Based on the definition of an inverse function, if we have two inverse functions f and g, then g of f ofx must be x. If we use the chain rule to differentiate, we can solve for frime of x in terms of g prime of f ofx. This is useful when you know the values for the original function but need to find information about its inverse and the inverse's derivative. Inverse trig functions which are different from the functions like cosecant and seccant have their own unique derivatives. You need to memorize these derivatives using the chart to the right. But notice that some of these derivatives are just opposite of each other. This makes memorizing the derivatives a bit easier. Higher order derivatives are successive derivatives of a function. There are a couple different notations used to indicate this. So, make sure that you're familiar with all of the notations shown. In motion problems, we track position, velocity, and acceleration. Velocity is the derivative of position and acceleration is the derivative of velocity or the second derivative of position. When the signs of velocity and acceleration are the same, the object is speeding up and when they're opposite, a particle is slowing down. Note that when velocity switches signs, the particle changes direction. Related rates problems involve finding the rate of change of one quantity in terms of another that is changing over time. We have a useful formula so that you can nail every related rates problem. First, write down what you know. Second, draw a picture if applicable. Third, write down an equation without derivatives that relates your variables and don't plug in any variables yet. And then fourth, implicitly differentiate the equation. Plug in your variables and solve. An example problem is shown to the right. Tangent line approximations are useful for estimating a function's values near a specific point. We use the fact that the derivative of a function is the slope of the tangent line and then apply point slope formula to approximate the function. For example, say we have a function f(x)= xubus x^2. We can try to approximate f of 5.1 without actually plugging 5.1 into the function. First we find the derivative of the function at x= 5 which turns out to be 100. Then we use point slope formula and we solve for y which turns out to be 106.5. Now comparing this to the actual value of 106.6 six. We see that the tangent line approximation is pretty accurate. So we want to take the limit as x approaches 0 of sinx /x. When we plug in x to take this limit, we get 0 / 0 which is something called an indeterminate form. Luckily we have low talls rule which says that when we get an indeterminate form, we can differentiate the numerator and denominator separately and then take the limit again. In this case, we find that the limit as x approaches zero of sinx /x equals 1. Sometimes you'll differentiate the numerator and denominator and you'll just get an indeterminant form. Again, if that happens, you just keep applying Li Tall's rule until you get a finite number or an infinite number. The mean value theorem says that if we have a function f that's continuous over the closed interval from a to b and differentiable over the open interval, then there must be a point where the instantaneous rate of change equals the average rate of change over that interval. Note that the mean value theorem doesn't tell us where this point is. We have to solve for it ourselves. For example, let's try to find where the instantaneous rate of change equals the average rate of change for f(x) = x^2 - 2x over the interval from 1 to 5. We find that the average rate of change equals 4. And then solving for the instantaneous rate of change, we get x= 3. And this is between our interval from 1 to 5, which checks out. Say we have a function f that's continuous over the open interval from a to b. The extreme value theorem guarantees that f has at least one minimum value and at least one maximum value on the closed interval from a to b. A global max or min is the highest or lowest value over a closed interval. While a local max or min is simply higher or lower than points directly to the left and right. A critical point is defined as any point where the derivative equals zero or the derivative is undefined. And all local extrema occur at critical points. When the first derivative is greater than zero, we know that f ofx is increasing. When the first derivative is negative, f ofx is decreasing. Whenever f ofx goes from increasing to decreasing, we know we have a local max. So therefore, when the derivative goes from positive to negative, we have a local max. And the opposite goes for the local min. Global extrema only occur at end points or critical points, but you need to check both. For example, say we want to find the global max of f(x)= xub - 3x^2 + 3 over this interval from -2 to 4. First, we check the end points and find that f of -2 is -17 and f of 4 is 19. And then we check the critical points which we get by solving um setting the first derivative equal to zero. After we compare all of the end points and critical points, we find that the global max is at x= 4 because f of 4 is 19, which is higher than all of the other end points and critical points. The second derivative of a function determines its concavity. When the second derivative is negative, that means that frime of x or the first derivative is decreasing and the function is concave down. On the other hand, when the second derivative is positive, that means the first derivative is increasing and the function is concave up. When f changes concavity or when the second derivative changes signs, we have a point of inflection. The second derivative test is another way to classify critical points. Whenever we have a local maximum, we know that the graph is going to be concave down. And whenever we have a local minimum, the graph's going to be concave up. So given x a critical point, if the second derivative is less than zero or the graph is concave down, we have a local max. And if the second derivative is greater than zero, we have a local min. The skills we've learned so far can be applied to optimization problems. Usually with an optimization problem, you'll get two equations. One that you're trying to maximize or minimize and the other that gives you a constraint on two variables. You use the constraint to get the first function in terms of one variable and then you can use the first derivative or second derivative test to find that function's extrema. From there, you find the global or local max or min and solve whatever the question is asking you for. Say we want to approximate the area under a function's graph. We can use something called a remon sum to find this. There are four different types of remon sums. Left remon sums use the function value as the top left corner of the rectangle. Right remon sums use the function value as the top right corner of the rectangle. The midpoint rule uses the function value as the midpoint of the top two corners of the rectangle. While the trapezoid rule instead of using rectangles uses trapezoids with the top left corner and top right corner each being unique points on the graph. As you can see in the picture to the right, the more rectangles you use in a remon sum, the more accurate the approximation is. Say we take the limit as the number of rectangles approaches infinity. When we do this, we get an integral. We can see that the integral is the limit as the width of the rectangles approaches zero of the sum of all the heights of the rectangles times the widths. There are two types of integrals. definite integrals and indefinite integrals. First, we'll talk about definite integrals. Definite integrals have bounds and the definite integral equals the area under the function between those two bounds. Definite integrals have properties like the properties of limits. There's constant multiplication, addition, and subtraction that can be pulled through the integral. Integrals have the unique properties that if you're integrating from A to C with B between A and C, that integral equals the integral from A to B plus the integral from B to C. Also, if you switch the order of the bounds, you negate the integral. Definite integrals give rise to the two fundamental theorems of calculus. One is that when you differentiate an integral, you just get the original function. Meaning that derivatives and integrals are inverses of each other. Note that the anti-derivative of a function f is the function that you differentiate to get f. And this gives rise to the other fundamental theorem of calculus. Integrating a function f from a to b equals the anti-derivative evaluated at b minus the anti-derivative evaluated at a. Next, let's talk about indefinite integrals. These integrals have no bounds and when you evaluate them, you get their anti-derivative plus a constant C. For example, let's take x^2. Think of what function you take the derivative of to get x^2. You take the derivative of 1/3 x cubed plus any constant c to get x^2. Sometimes indefinite integrals are a bit harder to evaluate though. Sometimes you have to use something called u substitution to simplify an integral and then solve it. First you find an expression in the integral to let u equal. Then you find the derivative du dx and you multiply both sides by dx to get a function for du. Then you substitute in u and change dx to du and then evaluate the integral. And then once you're done with that, you change u back to the function of x. To the right, we have an example problem showing you these steps. Another trick is integration by parts, which is only a BC topic. Remember back to the product rule that when you take a derivative of two functions, you get the derivative of the first time the second plus the first time the derivative of the second. Now integrate both sides and we get integral of u dv= uv minus the integral of v du. Look at the example problem to the right and pause for a walkthrough of integration by parts. Another trick for solving indefinite integrals is partial fractions. Say you have to integrate a rational function where the denominator is a higher magnitude than the numerator. You can split the rational function into two fractions. For example, we have 3x + 11 / x^2 - x - 6 gets split into 4x - 3 and 1/x + 2. And then you can easily integrate both these functions because they turn into ln functions. Another bc topic is improper integrals. These occur when one or both limits are unbounded. For example, the integral from 0 to infinity of 1 /x^2. What you do to solve these is you replace the unbounded limit with a variable and take the limit as that variable approaches infinity or negative infinity. Then you evaluate the integral like normal and then at the end you plug in that limit. Here we find that the integral from 0 to infinity of 1x^2 equals 0. Because this integral equals zero and not say infinity or negative infinity, this integral is convergent. If our integral turned out to be infinity or negative infinity, then the improper integral would be divergent. The first topic in differential equations is slope fields. Slope fields show the first derivative of a function at different points on the coordinate plane. When the slope is bigger, you depict this by using a longer arrow and you have to make sure that your arrows don't cross. Say you have a differential equation that you can't solve, but you want to approximate the solution using Oilers's method. What you do is you take a point, you find the derivative, and then you multiply the derivative by the change in x or the step size that you take. This is like the tangent line approximation and it gives you an approximate new value for y. Then you take this point, you multiply the derivative by the step size, and you repeat until you get to your desired x value. This gives you a pretty good approximation of the function's actual value. And the approximation is better the smaller the step size you use. Now for actually solving differential equations. When you get a problem asking for the particular solution of a differential equation, it means that you have to plug in some initial condition to get a specific function. When a problem asks for the general solution, you're usually leaving a plus C or multiplying by some constant that hasn't been solved for yet. There are four steps to solving differential equations. First you separate the variables, usually dy and dx or dy and dt. Then you integrate each side. You only need to add the constant of integration on the right side because you can just treat it as the constant on the right minus the constant on the left. After you do this, you solve for the dependent variable. And then if the question asks for a particular solution, you have to plug in your initial conditions to solve for that constant. Whenever you see a differential equation in the form dydt is proportional to y, it's going to be exponential. Specifically, if dydt equals ky, then y of t is going to be y0, some initial condition, * e kt. Logistic models can represent things like population growth where a function is bounded by a carrying capacity. They have the form dydt = k * 1 - l / y * y. Solving this differential equation yields y of t = l / 1 + b e kt where b is some constant that you can plug in an initial condition to solve for. Now for applications of integration, our first topic is average value. Definite integrals allow us to find the average value a of a function over an interval by a is 1 / b minus a of the integral from a to b of f ofx dx. Now it's time to talk more about position, velocity and acceleration. Remember that the derivative of position with respect to time is velocity and the derivative of velocity is acceleration. We can rewrite these relationships in terms of integrals to get the change in position or displacement is the integral of velocity over a time period. And the change in velocity is the integral of acceleration over a time period. To get total distance, we integrate speed, which is the absolute value of velocity. Say you want to find the area between two curves f ofx and g ofx. First you find the intersection points which will be the bounds of the integral. Then you get both functions in terms of the same variable either y or x. And then you integrate the top function minus the bottom function. Integrals can be used to find the volume of a solid generated by creating cross-sections with the function f. All the different cross-sections you can be asked to use are shown to the right. And you can also see the formulas used to solve for them. You can also use integrals to find the volume of a function that is revolved around the x or y axis. We use the equation v = pi * the integral from a to b of rx^2 where rx is the radius between the axis and the function. To get this volume, sometimes you want to revolve an area around an axis when you have space between the area and the axis. In this case, we use the washer method, where you subtract the inner radius squared from the outer radius squared. Our last topic for this unit is arclength, which gives us the total distance traveled by a curve in a bounded interval. To get the arclength, we integrate from a to b the square root of 1 + the derivative frime of x^2. Parametric functions give y and x separately in terms of t. We can get dydx in terms of our parametric equations by dividing dydt by dxdt. Assuming that dxdt does not equal zero because we can't divide by zero. We can also get the second derivative by taking the derivative with respect to t of dydx and then dividing that by dx dt. Again assuming that dxdt does not equal zero. Parametric equations give us another way to find the ark length. We can integrate the square root of dx dt ^2 plus dydt ^2 to find it. Parametric functions are really good for representing particle motion because the displacement vector can be modeled by x of t and y of t. And then if we just differentiate the components of this vector function, we get the velocity vector and the acceleration vector. Next up, we learn about polar coordinates and derivatives. Polar coordinates give a radius r as a function of theta where x= r cos theta and y = r sin theta. Like with parametric equations, we can get dydx in polar as dy d theta / dx d theta. Or if we don't want any y's or x's in our equation, it's frime of the * sin theta plus f of theta cos theta all / frime of theta cos theta minus f of theta sin theta. We can use the formula area= the integral from a to b of r^ 2 d theta where a and b are theta values and r is a function of theta to get the area bounded by a polar curve. If we want to find the area between two polar curves, we find the outer area and then we subtract the inner area. Our last unit of BC is about sequences and series. A sequence is a list of numbers while a series is a sum. Using limits, we can add an infinite amount of terms to form an infinite series. If this infinite sum equals positive or negative infinity, then we say the series diverges. While if the sum approaches a finite number, we say the series converges. We have different tests we can use to figure out if a series converges or diverges. Pause the video for a minute and review this table which has all the tests you need to know. A function f ofx can be approximated using a Mclloren polomial. The more terms we have in the polomial, the more accurate the approximation is. So if we take the limit as the number of terms approaches infinity, we call this a Mclloren series. Taylor series are a more general version of Mclloren series because Taylor series can be centered at any point C while Mclloren series are always centered about X=0. We want to know how accurate these approximations are though. We can use the lrangee errorbound formula to find the greatest possible error which is less than or equal to x - c to the n +1 / n +1 factorial time the maximum value of the n +1 derivative. Some functions can be modeled perfectly for all x values by their tailaylor series while other functions can only be modeled inside a certain interval of convergence. We can find the interval of convergence and the radius of convergence by running the ratio test on the tailaylor series with an example problem shown to the right. After we find this interval though, we need to double check the end points of the interval. Sometimes it's easier to differentiate or integrate a power series representing a function than it is to actually integrate or differentiate that function. We integrate or differentiate each term separately to find the derivative or integral of the function. The radius of convergence is always the same as the original function.