# If f(x)= tan8 x  and g(x) = e^(-3x ) , how do you differentiate f(g(x))  using the chain rule?

Jan 20, 2016

<$f ' \left(g \left(x\right)\right) = - 24 {e}^{- 3 x} . {\sec}^{2} \left(8 {e}^{- 3 x}\right)$

#### Explanation:

$f \left(g \left(x\right)\right) = f \left({e}^{- 3 x}\right) = \tan \left(8 {e}^{- 3 x}\right)$

using 'chain rule' to differentiate.

$f ' \left(g \left(x\right)\right) = {\sec}^{2} \left(8 {e}^{- 3 x}\right) \frac{d}{\mathrm{dx}} \left(8 {e}^{- 3 x}\right)$

$= {\sec}^{2} \left(8 {e}^{- 3 x}\right) . 8 {e}^{- 3 x} \frac{d}{\mathrm{dx}} \left(- 3 x\right)$

$= {\sec}^{2} \left(8 {e}^{- 3 x}\right) .8 {e}^{- 3 x} . \left(- 3\right)$

$\Rightarrow f ' \left(g \left(x\right)\right) = - 24 {e}^{- 3 x} . {\sec}^{2} \left(8 {e}^{- 3 x}\right)$