# How do you use the chain rule to differentiate y=cos^3(4x)?

Oct 28, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 12 {\left[\cos \left(4 x\right)\right]}^{2} \sin \left(4 x\right)$

#### Explanation:

What the chain rule is that it takes a composition of functions, and "peels" it back layer by layer. In the context of a calculation, you'd take the derivative of your outermost function leaving the inside as is, and then multiply by the derivative of the next outermost function, till you reach the end.

It's hard to describe properly in words, so let me show you with your example.

So, you have 3 different functions built into each other here. You have $f \left(x\right) = {x}^{3}$, $g \left(x\right) = \cos \left(x\right)$, and $h \left(x\right) = 4 x$.

Our outermost function is ${x}^{3}$. So we'd take the derivative of that while preserving the contents inside . This leaves:

$\implies 3 {\left[\cos \left(4 x\right)\right]}^{2}$

as our first term. Now for the next function, $\cos \left(x\right)$. Following the same principle:

$\implies 3 {\left[\cos \left(4 x\right)\right]}^{2} \cdot - \sin \left(4 x\right)$

Lastly, $4 x$:

$\implies 3 {\left[\cos \left(4 x\right)\right]}^{2} \cdot - \sin \left(4 x\right) \cdot 4$

Cleaning this up gives:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 12 {\left[\cos \left(4 x\right)\right]}^{2} \sin \left(4 x\right)$