# How do you differentiate f(x)=(cos x+ 4sin^2x^2)^6 using the chain rule?

##### 1 Answer
Jun 2, 2018

$6 {\left(\cos x + 4 {\sin}^{2} {x}^{2}\right)}^{5} \left(- \sin x + 16 x \sin {x}^{2} \cos {x}^{2}\right)$

#### Explanation:

In several stages.

First stage - chain rule for the bracket as a whole:
$\frac{d}{\mathrm{dx}} \left[{\left(\cos x + 4 {\sin}^{2} {x}^{2}\right)}^{6}\right]$
$=$
$6 {\left(\cos x + 4 {\sin}^{2} {x}^{2}\right)}^{5} \cdot \frac{d}{\mathrm{dx}} \left[\cos x + 4 {\sin}^{2} {x}^{2}\right]$

Second stage - two terms, first one simple, second chain rule again:
$\frac{d}{\mathrm{dx}} \left[\cos x\right] = - \sin x$
$\frac{d}{\mathrm{dx}} \left[4 {\sin}^{2} {x}^{2}\right] = 8 \sin {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left[\sin {x}^{2}\right]$

Third stage - chain rule again
$\frac{d}{\mathrm{dx}} \left[\sin {x}^{2}\right] = \cos {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left[{x}^{2}\right]$

Fourth stage - simple differentiation
$\frac{d}{\mathrm{dx}} \left[{x}^{2}\right] = 2 x$

Work back through the process:
$\frac{d}{\mathrm{dx}} \left[{x}^{2}\right] = 2 x$
$\frac{d}{\mathrm{dx}} \left[\sin {x}^{2}\right] = 2 x \cos {x}^{2}$
$\frac{d}{\mathrm{dx}} \left[4 {\sin}^{2} {x}^{2}\right] = 16 x \sin {x}^{2} \cos {x}^{2}$

$\frac{d}{\mathrm{dx}} \left[{\left(\cos x + 4 {\sin}^{2} {x}^{2}\right)}^{6}\right]$
$=$
$6 {\left(\cos x + 4 {\sin}^{2} {x}^{2}\right)}^{5} \left(- \sin x + 16 x \sin {x}^{2} \cos {x}^{2}\right)$