# If f(x)= cot2 x  and g(x) = e^(1 - 4x ) , how do you differentiate f(g(x))  using the chain rule?

##### 1 Answer
Oct 13, 2016

$\frac{8 {e}^{1 - 4 x}}{\sin} ^ 2 \left(2 {e}^{1 - 4 x}\right)$ or $8 {e}^{1 - 4 x} {\csc}^{2} \left(2 e \left(1 - 4 x\right)\right)$

#### Explanation:

$f \left(g \left(x\right)\right) = \cot 2 {e}^{1 - 4 x}$

Let $g \left(x\right) = u$

$f ' \left(u\right) = \frac{d}{\mathrm{du}} \cot 2 u = \frac{d}{\mathrm{du}} \frac{\cos 2 u}{\sin 2 u} = \frac{- 2 \sin \left(2 u\right) \sin \left(2 u\right) - 2 \cos \left(2 u\right) \cos \left(2 u\right)}{\sin} ^ 2 \left(2 u\right)$
$= \frac{- 2 {\sin}^{2} \left(2 u\right) - 2 {\cos}^{2} \left(2 u\right)}{\sin} ^ 2 \left(2 u\right)$
$= - \frac{2}{\sin} ^ 2 \left(2 u\right)$

$g ' \left(x\right) = - 4 {e}^{1 - 4 x}$

Using chain rule: $f ' \left(g \left(x\right)\right) = f ' \left(u\right) \cdot g ' \left(x\right)$

$= - \frac{2}{\sin} ^ 2 \left(2 u\right) \cdot - 4 {e}^{1 - 4 x}$

$= - \frac{2}{\sin} ^ 2 \left(2 {e}^{1 - 4 x}\right) \cdot - 4 {e}^{1 - 4 x}$

$= \frac{8 {e}^{1 - 4 x}}{\sin} ^ 2 \left(2 {e}^{1 - 4 x}\right)$ or $8 {e}^{1 - 4 x} {\csc}^{2} \left(2 e \left(1 - 4 x\right)\right)$