# How do you differentiate f(x)=e^tan(2-x^3)  using the chain rule?

Nov 6, 2017

$f ' \left(x\right) = - \frac{3 {x}^{2} {\sec}^{2} \left({x}^{3} - 2\right)}{e} ^ \left(\tan \left({x}^{3} - 2\right)\right)$

#### Explanation:

We have $f \left(x\right) = {e}^{\tan \left(2 - {x}^{3}\right)}$. We can see that $f$ is a composition of three different functions, so in order to find $f '$, we need to differentiate each of these three functions as if their arguments were a single variable and then multiply their derivatives.

$\frac{d}{\mathrm{dx}} {e}^{u} = {e}^{u}$

$\frac{d}{\mathrm{dx}} \tan v = {\sec}^{2} v$

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

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

We have the derivative, but it doesn't look particularly nice, so we can rewrite it to look neater.

$\sec \left(- \theta\right) = \sec \theta \Rightarrow \sec \left(2 - {x}^{3}\right) = \sec \left({x}^{3} - 2\right)$

$\tan \left(- \theta\right) = - \tan \theta \Rightarrow \tan \left(2 - {x}^{3}\right) = - \tan \left({x}^{3} - 2\right)$

${e}^{- u} = \frac{1}{e} ^ u \Rightarrow {e}^{\tan} \left(2 - {x}^{3}\right) = {e}^{- \tan \left({x}^{3} - 2\right)} = \frac{1}{e} ^ \tan \left({x}^{3} - 2\right)$

$\therefore f ' \left(x\right) = - \frac{3 {x}^{2} {\sec}^{2} \left({x}^{3} - 2\right)}{e} ^ \left(\tan \left({x}^{3} - 2\right)\right)$