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### Course: AP®︎ Calculus AB (2017 edition)>Unit 3

Lesson 9: Chain rule

# Worked example: Derivative of cos³(x) using the chain rule

Now let's tackle a worked example of composite functions, using the example f(x) = cos³(x). By applying the chain rule, we differentiate this function, breaking it down into its components x³ and cos(x). This helps us understand how to handle other complex derivatives with ease.

## Want to join the conversation?

• What does 'grok' mean? ,
• Robert Heinlein fist used the term "grok" in his Novel Stranger in a Strange Land. "Grok" essentially means to have a deep, comprehensive, thorough understanding of a concept. Perhaps the "grok" in this context would mean that by now it should be second nature.
• how did sal multiply 3cosx by the -sinx
• You can see this expression this way: 3 * cos(x) * -1 * sin(x). And remember, multiplication is commutative, so you can just multiply the terms in front of the functions and you end up with -3 * cos(x) * sin(x).
• I don't understand why dv/du is the same thing as v'( u(x) ).
• Think of the dy/dx notation as a function; dy is the thing you are differentiating, and dx is what you are differentiating for (in this case u(x))
• Albeit I should see past lessons probably, (I am missing some basic knowledge) but why does cos become -sin?
• Here's a link to some common derivatives that includes cos and sin.
• Why is the dv in "dv/du" equal to d(cosx^3) even though the function v in v(x) is x^3, not cosx^3?
• Perhaps Sal should have written it a little differently. it is more ccurate to say
dv(u(x)) / du(x) * du(x) / dx So the derivative of v(u(x)) with respect to u(x) times the derivative of u(x) with respect to x.

Let me know if that doesn't make sense.
• Can't we just think like every differential rule needs chain rule (for the concept)? it's just at some point you'd reach f(x) = x, and since f'(x) = 1 you can just cross it off in multiplications.
I always think of it that way (similar to pythagorean theorem is actually law of cosines, but the "2abcos(theta)" always results in substraction by zero)
• Yes — and this will be a very helpful way to think of derivatives later on when you get to implicit differentiation!
• So when Sal is taking the derivative of cosx^3 with respect to cosx instead of x, would that be like graphically taking the derivative but on a graph with (cosx) and y axes instead of just x and y axes?
• Yes, every x value would then be a cos x value, resulting in a (cos x) axis.
• @ Sal multiplies for the final answer. Why does the (cosx)^2 become cos^2x? Is the x no longer squared? Thanks!
• sfrazee5001,

Ah, I see. There are really two ways of writing. If you look at the upper left of the original f(x) function, we see f(x) = cos^3(x) = (cos x)^3. They really mean the same thing; it's just two different ways of writing the same thing.

However, be careful about negative powers. For example, sin^-1(x) is not the same as (sin x)^-1 or csc x.
• Isn't V basically equal to x^3 ?!

If I'm right, then why does dV/dU = d(cosx)^3/dcosx, and not dx^3/dcosx?

Thanks a lot,
• Actually, df/du=v'( u(x) ), there is a slight error at .
f'(x) should be equal to df/du*du/dx , not dv/du*du/dx. Very well spotted.
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