What is the derivative of cos^2(x^3)?

Jan 22, 2016

$f ' \left(x\right) = - 6 {x}^{2} \cos \left({x}^{3}\right) \sin \left({x}^{3}\right)$

Explanation:

$f \left(x\right) = {\cos}^{2} \left({x}^{3}\right)$

Let's break your function down as a chain of functions:

$f \left(x\right) = {\left[\textcolor{b l u e}{\cos \left({x}^{3}\right)}\right]}^{2} = {\textcolor{b l u e}{u}}^{2}$

where

$u = \cos \left(\textcolor{v i o \le t}{{x}^{3}}\right) = \cos \left(\textcolor{v i o \le t}{v}\right)$

where

$v = {x}^{3}$

Thus, the derivative of $f \left(x\right)$ is:

$f ' \left(x\right) = \left[{u}^{2}\right] ' \cdot u ' = \left[{u}^{2}\right] ' \cdot \left[\cos v\right] ' \cdot v '$

Now, let's compute those three derivatives:

$\left[{u}^{2}\right] ' = 2 u = 2 \cos {x}^{3}$

$\left[\cos v\right] ' = - \sin v = - \sin {x}^{3}$

$\left[v\right] ' = \left[{x}^{3}\right] ' = 3 {x}^{2}$

Thus, you can compute your derivative as follows:

$f ' \left(x\right) = \left[{u}^{2}\right] ' \cdot \left[\cos v\right] ' \cdot v '$

$= 2 \cos {x}^{3} \cdot \left(- \sin {x}^{3}\right) \cdot 3 {x}^{2}$

$= - 6 {x}^{2} \cos \left({x}^{3}\right) \sin \left({x}^{3}\right)$