So this video should help you review the derivative rules you would have learned in one of your first units in a high school calculus course. So let's take a look at the derivative rules you would have learned and then just do some quick examples that use each of those rules. So I'm going to start off just by showing you this list. I'm not going to talk about it right now.
I'll reference them one by one as we use them. But here's a list of all the derivative rules that you would have used to start off your calculus course. So we got power rule, constant multiple rule, sum rule, difference rule, product rule, quotient rule, and so on. rule, power of a function rule, chain rule.
So you know go back, pause the video if you want to take a close look at these now, but I'll reference them as we go. So this first example here, I've got this function negative x to the 5 plus 5x cubed plus cube root of x squared. If I want to take the derivative of this function, what I'd have to do is I'd have to use, you know, the power rule and constant multiple rule. So if I want the derivative, I'll reference the rule over here right now. So if I want to take the derivative of c times x to the n, it would equal, I would find the product of the exponent and the coefficient, you know, so I'd have c times n multiplied by the power raised to an exponent one less to what it was originally.
So we can use that rule to differentiate each of these three terms. I'm going to start maybe by rewriting this cube root of x squared. So I'm not going to take the derivative yet.
Oh, and I should mention for each of the functions that I give you in these questions, we're going to be finding the derivative of each of them. So I'm just going to, each example, I'm just going to show you a function and underneath we're going to find this derivative. Before I find the derivative, I'm going to rewrite that radical as a power with a rational exponent.
So cubed root of x squared, that's the same. thing as saying x to the power of 2 thirds. And now I can use this rule, this power rule combined with the constant multiple rule, to find its derivative.
So now I'll find f prime of x. So I just have to do 5 times negative 1, that's negative 5x, and then reduce the exponent by 1. Negative 5x to the 4 is the derivative of negative x to the 5. Now differentiate 5x cubed with respect to x. I would get 5 times 3, which is 15x squared.
And this one, I would bring the 2 thirds down, write it as the coefficient, x to the 2 thirds minus 1. So that's 2 thirds minus 3 thirds. That would be x to the negative 1 third. Right, 2 thirds minus 3 thirds, negative 1 third. And you shouldn't leave an answer with a negative exponent. So you would take that power of x and write it in the denominator with 3, and that would change the sign of the exponent.
So f prime of x equals negative 5x to the 4 plus 15x squared. plus 2 over 3x to the third, or you could write that as 3 cubed root x, whatever you'd rather. There we go, there's our first derivative. After you learned the power rule for derivatives, you probably went on and learned the product rule of derivatives. If we're doing the derivative of a product of two functions, f at x and g at x, the derivative would equal the derivative of the first function times the second function plus the derivative of the second function times the first function.
So let's see how that rule works. I want to find the the derivative of the product of that function and that function, it would equal The derivative of the first function, derivative of 11x plus 2 is 11, times the second function, negative 5 plus 3x squared, plus the derivative of the second function, so the derivative of this function, well the derivative of a constant is 0, but the derivative of 3x squared would be 6x, so 6x times 11x plus 2. and then we would just expand this out and collect any like terms we would have. So distribute the 11 into this binomial, the 6x into this binomial, and then collect our like terms.
So I'll just do that quickly, expand the 11 in to the first binomial, and now expand the 6 in to the second binomial, and then collect any like terms we have. Well these are like terms, they both have an x squared, so we can put those together. and we would get 99x squared plus 12x minus 55. That would be our derivative. Let's try the next one.
Let's use quotient rule. So after product rule, we probably learned quotient rule. We want the derivative of a quotient of two functions.
That would equal the derivative of the top function times the bottom function minus the derivative of the bottom function times the top function all over the bottom function squared. So what does that look like for this one? I would need to do the derivative of the top function, so the derivative of x squared minus 4. Well, the derivative of x squared is 2x, and the derivative of negative 4 is 0, so the derivative of the top is just 2x, times the bottom function, minus the derivative of the bottom function, which would just be 2, times the top function, all over the bottom function squared.
And then we can just expand the numerator. collect. So if I expand the numerator I'd have 4x squared plus 10x, that's me expanding this 2x into both of these terms, and now I'm going to distribute the negative 2 to both these terms, so minus 2x squared plus 8, all over 2x plus 5 squared.
And my final line, 4x squared minus 2x squared is 2x squared. So I'm just collecting my like terms. Plus 10x plus 8 over 2x plus 5 squared.
Now you can check if the numerator factors. We can put it into factored form if we want, but that's not going to help it reduce with anything in the denominator, so I can leave it in standard form. Or you can try and factor it. It's totally up to you.
So if we wanted to factor it, what I'm saying is, you know, we could... common factor of 2 first off, and this actually factors further, right? This would just be multiplies to 4, adds to 5, or 4 and 1, so it could be 2x plus 4x plus 1 over the denominator, but that doesn't help us reduce anything with the denominator, so either answer would be fine for that. Alrighty, let's do a chain rule question now. So this one we're going to have to use a combination of our rules.
So if we have, if we want to do the derivative of a power where the base of the power is itself a function that can be differentiated, what we have to do is do the derivative of the outside function with respect to the inside function. So, you know, use the power rule, bring this down, write it as the coefficient and reduce the exponent by one. It's now the coefficient, reduce the exponent by one. and leave the inside function the way it is, leave the base of the power the way it is, but then multiply that by whatever the derivative of the base of the power was.
So in this case, I would differentiate the outside function with respect to the inside function, so write the 2 as the coefficient of the power, leave the base of the power as it is, and reduce the exponent by 1, and then I need to multiply this by the derivative of the base of the power. So the derivative of 3x minus 2 is 3. So here I have, well, 2 times 3 is 6, so I have 6 times 3x minus 2. I could expand that further if I want, or I could leave it in factored form. Either would be fine.
Okay, so now let's do some questions where we have to use all of these rules together. So with this one, I'm going to have to use a combination of chain rule and product rule. I'm doing product rule.
I have a function times a function. For each of those functions, I would have to use chain rule. So if I want the derivative of this, So product rule tells me to do the derivative of the first function, so that'd be 3 2x minus 3 squared times whatever the derivative of the inside function is, which would be 2, times the second function, 3x minus 1 squared, plus derivative of the second function times the first function. So derivative of this second function would be 2 3x minus 1 to the 1 times the derivative of 3x minus 1, which is 3. multiplied by the first function 2x minus 3 cubed. Now we could expand this if we want to leave it in, if we want to put it into standard form, you know, fully expanded and simplified form, or we can put it into factored form.
And that's actually going to be easier because if you notice we have two terms here, right? This plus sign separates into two terms. And what's the same between these two terms?
Well, they both have a 3 times 2, which is 6. This one has 2 times 3, which is 6. So I could common factor out a 6. from both terms. They both have a 2x minus 3 factor. This one in fact has two 2x minus 3 factors.
This one has three 2x minus 3 factors. So always take out the one with the lower exponent, so I can take out a 2x minus 3 squared from both terms. And they both have a 3x minus 1 factor.
This one has two of them. This one just has one of them. So I can take out the one with the lower exponent. I could take out a 3x minus 1 to the 1. And now after I've taken it out, after I'm common factoring it out from both terms, I need to divide both of the terms by what I took out. So let's look just at this first term here.
Let's look at this term here. I'm going to divide that term by what I took out. So 6 divided by 6 is 1. 2x minus 3 squared divided by 2x minus 3 squared is 1, and 3x minus 1 squared divided by 3x minus 1 is 3x minus 1. Plus, now let's divide the second term here, let's divide this term by what we took out.
So I've got 6 divided by 6, that's 1, 3x, or I'll do 2x minus 3 first. 2x minus 3 cubed divided by 2x minus 3 squared is 2x minus 3. And then 3x minus 1 divided by 3x minus 1 is 1, so I don't have to write that. So there we go.
It's common factored. We should get put in fully simplified factored form, so I should simplify this factor that I've created here. I've got 6, 2x minus 3 squared, 3x minus 1. And then this factor here, I've got 3x plus 2x, that's 5x. And then negative 1 plus negative 3, that's negative 4. There we go, there's the derivative in fully factored form, right? It's just the product of these things here.
6, 2x minus 3 squared, 3x minus 1, 5x minus 4. Alright, let's try one more. So this one we're going to have to use quotient rule and chain rule. So I would need to do the derivative of the numerator times the denominator, right?
Derivative of top, which would be 24x squared, times the denominator. And I'll write it with a rational exponent, right? root of 3x minus 2 is 3x minus 2 to the half minus derivative of the bottom so when i do the derivative of this i'd have to use chain rule right bring the half in front keep the base of the power the same subtract one from the exponent half minus one is negative a half times the derivative of the base so that would just be three times the top function 8x cubed and this would be all over my bottom function squared. Okay, let's do a little bit of simplifying here.
First off, let's look at, in the numerator, I've got a 3x minus 2 factor with my first term, and a 3x minus 2 factor in my second term. So I can common factor it out, but you always common factor out the one with the lower exponent. So in this case, Negative 1 half is the lower exponent.
So let's take out a 3x minus 2 to the negative 1 half. So I'm going to take out 3x minus 2 to the power of negative 1 half, and then divide both of the terms by what I took out. So the 24x squared in the first term will stay, and when I do 3x minus 2 to the half divided by 3x minus 2 to the negative 1 half, I get 3x minus 2 to the 1, or just 3x minus 2. Minus. The second term.
I've got this half times 3 times 8. You know, that's just going to be 12, right? I've got, well, half of 8 is 4 times 3 is 12, so that's 12. I took out the 3x minus 2 to the negative half exactly, so that's gone. And I've got that x cubed.
And this is all over 3x minus 2. It's just gonna be to the power of 1 now, right? Because exponent on top of exponent, they multiply together, 2 times is 1. So I just have 3x-2. What I'm gonna do next, I'm gonna simplify what's this factor right here by expanding and collecting, but also I'm gonna bring this power to the denominator to change the sign of its exponent to become positive.
So we'll do that all in one step here. So when I expand here, I've got 72x cubed minus 48x squared minus 12x cubed over 3x minus 2 now to the positive 1 half, and another factor of 3x minus 2 that was already there. So I can combine those powers that are in the denominator, because they have the same base, so I can add their exponents. 1 half plus 1 is 1 and 1 half, or 3 over 2, written as a fraction.
And numerate, so I'll do that denominator, sorry, 3x minus 2 to 1 and a half, or 3 over 2. Numerator, I can collect the 72x squared and the 48x squared together. So I want to check and make sure I did that properly. 72x, oh, it's 72x cubed. I don't know why I wrote squared.
It's 72x cubed. So up here when I expanded the 24x squared times 3x, the x's are being multiplied, so of course I should have added the exponents 2 and 1 together, that's 3, so 72x cubed. That way it collects with the 12x cubed, so 72x cubed minus 12x cubed is 60x cubed.
minus 48x squared. Now that numerator, you can factor it, it's just it's not going to help reduce with anything in the denominator. Like I could take out a 12x squared from both of those terms if I wanted to, and it would give me 5x minus 4. So if you want it in fully factored form, there you go. If you want it expanded and simplified, there you go. Either would be fine.
Okay, last question of this review I'm going to do with you. This question, I just want to make sure you remember what the equation of these derivatives we're finding actually represent, and how we can use them to find equations of tangent lines. So find an equation for the tangent at x equals 1 to the curve y equals 2x over x plus 1 to the 6th.
So keep in mind, when we find the equation of a derivative, all of these derivatives that we've been finding, they tell you the instantaneous rate of change, or the slope of the tangent, at any value of x we want to the original function. So if we want the equation of the tangent at x equals 1 to this curve, well let me show you graphically what I'm talking about. So here's the curve 2x over x plus 1 to the 6, it's this red function here, and what we want to do is we want to figure out the equation of a tangent line to this point on the function when x is 1. So a tangent line is a line that touches the function at exactly only that one spot, so it's going to look like this.
We want the equation of the line that I was kind of outlining there. So first thing, if we want the equation of a line, well we need to know a point on the line and we need to know its slope. So finding a point on the line should be relatively simple.
We would just need to go ahead and evaluate, well what's the value of this function when x is 1? So let's find the value of y when x is 1. So we would just do 2 times 1 over 1 plus 1, all to the power of 6. So you know, we've got 2 over 2, which is 1, to the power of 6. and that's just one. So we have a point, the point 1, 1 is our x, y point.
So that point right there, if I graph the point 1, 1, it should be right on top of that blue point that you see. There we go, right on top of it. So we have the coordinates of that point.
Now we need the slope of the tangent line that will touch the function at that one point. So if we want the slope of the tangent line, we'll have to use the derivative. So let me rewrite the function.
It was 2x over x plus 1 to the sixth. So let's find y prime. Let's find the derivative of this function.
So y prime equals, well, it's a power where the base of the function is differentiable. So we should use our chain rule here by doing the derivative of the outside function with respect to the inside function. So keep the base of the power the same, but use your power rule. Reduce the exponent by 1. Now this needs to be multiplied by the derivative. of the base of the power.
So the derivative of 2x over x plus 1, well we're actually going to have to use quotient rule for that. So it'll be the derivative of the top, which is 2, times the bottom, minus the derivative of the bottom, times the top, all over, sorry let me write that again, all over the bottom function squared. So let's simplify this a little bit. I mean, we could just plug 1 into this equation now if we wanted to. I think it would be useful to simplify it a little bit.
So we have 6, and then this power of 5 goes on the 2x and the x plus 1. So 2x to the power of 5, well, 2 to the power of 5, 2, 4, 8, 16, 32. So we've got 32x, oh, I won't write the bracket, I'll write it like this. So we have... 32x to the 5, right? 2 to the 5 is 32x to the 5 over x plus 1 to the 5. And then the numerator here, I would have 2x plus 2 minus 2x.
So 2x plus 2 minus 2x, that's just 2 over x plus 1 squared. And when we multiply this out, you know, we're going to have 6 times 32 times 2. which is 384. So I've got 384 x to the 5, and then multiply the denominators. Well, it's two powers of the same base.
Add the exponents, I get 7. Now let's find y prime at 1, right? This equation will tell us the slope of the tangent at any point we want on the original function, and we want to figure out the slope of the tangent when x is 1. So let's find y prime at 1. Now like I said, we could have plugged 1 into any version of this derivative. It didn't have to be the fully simplified version. We could have plugged in enough here. But I think it was a useful exercise just to practice simplifying.
So we've got... Here we go. So y prime at 1, we've got 384 over 2 to the 7, and we get an answer of 3. That means the slope of the tangent is 3. Our slope m is 3. So our equation of our tangent line, we'll write it in y equals mx plus b form, and we know our point, our xy point, was 1, 1, so I can replace the x and the y with ones, and our slope we just found out was three, now let's just solve for the y-intercept, and when we rearrange this we'll get two. So the equation of our line is y equals 3x minus two, and if I graph that just to make sure you understand what we're doing here, y equals 3x minus two, notice that the line is tangent to the function, meaning it only touches the function at that one exact point.
So there we go. So there's the main derivative rules you would have learned at the beginning of your high school calculus course. Hopefully that was a good review.
If you want more extensive review on each of those individual topics, make sure you go back and watch the videos I posted on each individual topic. Alright, hope that helped.