# If f(x)= sin6 x  and g(x) = e^(3+2x ) , how do you differentiate f(g(x))  using the chain rule?

Jan 6, 2016

Step by step explanation is given below.

#### Explanation:

Chain rule $\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \ast g ' \left(x\right)$

$f \left(x\right) = \sin \left(6 x\right)$
Differentiating with respect to x using chain rule

$f ' \left(x\right) = \cos \left(6 x\right) \frac{d}{\mathrm{dx}} \left(6 x\right)$
$f ' \left(x\right) = 6 \cos \left(6 x\right)$

$g \left(x\right) = {e}^{3 + 2 x}$
Differentiating with respect to x using chain rule
$g ' \left(x\right) = {e}^{3 + 2 x} \frac{d}{\mathrm{dx}} \left(3 + 2 x\right)$
$g ' \left(x\right) = {e}^{3 + 2 x} \left(2\right)$
$g ' \left(x\right) = 2 {e}^{3 + 2 x}$

We are to find derivative of $f \left(g \left(x\right)\right)$

$f ' \left(g \left(x\right)\right) = 6 \cos \left(6 {e}^{3 + 2 x}\right)$

Using the chain rule
$\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \ast g ' \left(x\right)$

$\left(f \left(g \left(x\right)\right)\right) ' = 6 \cos \left(6 {e}^{3 + 2 x}\right) \left(2 {e}^{3 + 2 x}\right)$

$\left(f \left(g \left(x\right)\right)\right) ' = 12 {e}^{3 + 2 x} \cos \left(6 {e}^{3 + 2 x}\right)$ Answer