Manipulating functions before differentiation | Derivative rules | AP Calculus AB | Khan Academy

What I have listed here is several of the derivative rules that we've used in previous videos. If these things look unfamiliar to you, I encourage you maybe to not watch this video because in this video we're going to think about when do we apply these rules, what strategies, and can we algebraically convert expressions so that we can use a simpler rule.

But just as a quick review, this is of course the power rule right over here—very handy for taking derivatives of X raised to some power. It's also can use that with the derivative properties of sums of derivatives or differences of derivatives to take derivatives of polynomials. This right over here is the product rule. If I have an expression that I want to take the derivative of and I can think of it as the product of two functions, well then the derivative is going to be the derivative of the first function times the second function plus the first function times the derivative of the second.

Once again, if this looks completely unfamiliar to you or you're a little shaky, go watch the videos, do the practice on the power rule or the product rule, or in this case the quotient rule. The quotient rule is a little bit more involved, and we have practice in videos on that. I always have mixed feelings about it because if you don't remember the quotient rule, you can usually—or you can always—convert a quotient into a product by expressing this thing at the bottom as F of x or by expressing this as F of x * G of x to the 1. So you could take the derivative with a combination of the products.

And this fourth rule over here is the chain rule. If any of this is looking unfamiliar, again don't watch this video. This video is for folks who are familiar with each of these derivative rules or derivative techniques and now want to think about, well, what are strategies for deciding when to apply which.

So let's do that. Let's say that I have the expression—let's say I'm interested in taking the derivative of (x^2 + x - 2) over (x - 1). Which of these rules or techniques would you use? Well, you might immediately say, "Hey look, this looks like a rational expression. I could say look, I could say this is my F of x right over here. I could say this is my G of x right over here," and I could apply the quotient rule. This looks like the quotient of two expressions, and you could do that. If you do all the mathematics correctly, you will get the correct answer.

But in this case, it's good to just take a little time to realize, well, can I simplify this algebraically? So maybe I can do a little bit less work. If you look at it that way, you might realize, "Wait, what if I factored this numerator? I can factor it as (x + 2) * (x - 1), and then I could cancel these two factors out."

I can say, "Hey, you know what? This is going to be the same thing as the derivative with respect to X of (x + 2)," which is much, much, much, much more straightforward than trying to apply the quotient rule here. You would just take the derivative with respect to X of X, which is just going to be one, and the derivative with respect to X of two is just going to be zero. So all of this is just going to simplify to one.

For taking the derivative of that, you're essentially just using the power rule. Once again, just a simple algebraic recognition, things become much, much, much more simple. Let's do another example. So let's say that you were to see, or someone were to ask you to take the derivative with respect to X of, let me see, so let's say you had (x^2 + 2x - 5) over X.

Once again, you might be tempted to use the quotient rule. This looks like the quotient of two expressions, but then you might realize, well, there's some algebraic manipulations I can do to make this simpler. You could express this as a product. You could say that this is the same thing as—and I'm just going to focus on what's inside the parentheses or inside the brackets—this is the same thing as X to the 1 * (x^2 + 2x - 5).

Then you might want to apply the product rule, but there's even a better simplification here. You could just divide each of these terms by X, or one way to think about it—distribute this (1/X) across all the terms. X to the 1 is the same thing as (1/X), and if you do that, X²/X is going to be X, 2x/X is going to be 2, and then -5/X, well you could write that as -5 over X or -5x^(-1).

Now taking the derivative of this with respect to X is much easier than using either the quotient or the power rule. This is going to be, let's see, the derivative of that is just going to be one, the derivative of two is just going to be zero, and here, even though you have a negative exponent, it might look a little intimidating, this is just taken using the power rule. So -1 * -5 is +5, X to the—if we take one less than -1, we're going to go to the -2 power.

So once again, making this algebraic recognition simplified things a good bit. Let's do a few more examples of just starting to recognize when we might be able to simplify things to do things a little bit easier. So let's say that someone said, "Hey, take the derivative with respect to X," and I'm using X as our variable that we're taking the derivative with respect to, but obviously this works for any variables that we are using.

So let's say we're saying (sqrt{X}) over (X²) – pause this video and think about how would you approach this if you want to take the derivative with respect to X of (sqrt{X}) over (X²). Well, once again, you might say this is a quotient of two expressions—you might try to apply the quotient rule—or you might recognize, well look, this is the same thing—let me just focus on what's inside the brackets. You could view this as X^(-2) * (sqrt{X}) * (sqrt{X}), and then you might want to use a product rule. But you could simplify this even better.

You could say this is the same thing as X^(-2) * X^(1/2) and now, just using our exponent properties, -2 + 1/2 is -3/2. So this is the derivative of X^(-3/2). Here once again, we took something that we thought we might have to use a quotient rule or use the product rule, and now this just becomes straightforward using the power rule.

So this is just going to be equal to—bring the -3/2 out front—(-3/2) * X^(-3/2 - 1), which is X^(-5/2). So once again, just before you apply the quotient rule—and sometimes even the product rule—just see if there is an algebraic simplification, sometimes a trigonometric simplification that you can make that eases your job, that makes things less hairy.

As a general tip, I can't say this is going to be always true, but if you're taking some type of exam and you're going down some really hairy route—which the quotient rule will often take you—it's a good sign that, hey, take a pause before trying to run through all of that algebra to apply the quotient rule, and see if you can simplify things.

So let's give another example, and this one there's not an obvious way, and it really depends on what folks' preferences are. But let's say you want to take the derivative with respect to X of 1 over (2x^5 - 5). Sorry, 1 over (2x - 5), I should say. Well here, you could immediately apply the quotient rule here for the numerator. You could view that as F of x; you could view this as the same thing as the derivative with respect to X of (2x - 5) to the -1 power.

Now, in this situation, you would use a combination of the power rule and the chain rule. You'd say, "Okay, my G of X is (2x - 5), and F of G of x is going to be this whole expression." If you applied the chain rule, this is going to be the derivative of the outside function—our f of x with respect to the inside function—or the derivative of f of G of X with respect to G of X. So it's going to be negative—bring that negative out front—so essentially, just going to use the power rule here: -1 * (2x - 5)^(-2).

Then we multiply that times the derivative of the inside function. The inside function's derivative—the derivative of (2x) is 2; the derivative of (-5) is 0, so it's going to be times 2. Of course, you can simplify it to -2 * all of this. Let me do one more example here just to hit the point home, and once again there isn't a must way there isn't a way that you have to do this, but just to let you appreciate that there are multiple ways to approach these types of derivatives.

So let's say someone said take the derivative of (2x + 1)^2. Pause the video and think about how you would do that. Well, one way to do it is just to apply the chain rule, just like we just did. So you could say, "Alright, it's going to be the derivative of the outside with respect to the inside," so it's going to be 2 * (2x + 1) to the first power—taking one less than that—times the derivative of the inside, which is just going to be two. This is going to be equal to 4 * (2x + 1), which is equal to—we could distribute the four—we could say it's 8x + 4.

That's a completely legitimate way of doing it. Now, there are other ways of doing it. You could expand out (2x + 1)^2; you could say, "Hey, this is the same thing as the derivative with respect to X of (4x^2 + 4x + 1)," and now you would just apply the power rule. So a little bit of extra algebra up front, but you can just go straightforward with the power rule, and you're going to get this exact same thing.

So the whole takeaway here is pause, look at your expression, see if there's a way to simplify it, and it's especially a good thing if you can get out of using the quotient rule because that sometimes is just hard to know or remember. Even when you do remember, it can get quite hairy quite fast.