# How do you find the derivative of arctan(e^x)?

Dec 22, 2016

$\frac{d}{\mathrm{dx}} \arctan \left({e}^{x}\right) = \frac{{e}^{x}}{{e}^{2 x} + 1}$

#### Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you can use the chain rule.

$y = \arctan \left({e}^{x}\right) \iff \tan y = {e}^{x}$

Differentiate Implicitly:

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x}$ ..... [1]

Using the $\tan \text{/} \sec$ identity;

${\tan}^{2} y + 1 \equiv {\sec}^{2} y$
$\therefore {\left({e}^{x}\right)}^{2} + 1 = {\sec}^{2} y$
$\therefore {e}^{2 x} + 1 = {\sec}^{2} y$

Substituting into [1]

$\therefore \left({e}^{2 x} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x}}{{e}^{2 x} + 1}$

Dec 22, 2016

$\frac{d}{\mathrm{dx}} \arctan \left({e}^{x}\right) = {e}^{x} / \left({e}^{2 x} + 1\right)$

#### Explanation:

Using implicit differentiation together with the known derivatives $\frac{d}{\mathrm{dx}} \tan \left(x\right) = {\sec}^{2} \left(x\right)$ and $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$,

let $y = \arctan \left({e}^{x}\right)$

$\implies \tan \left(y\right) = \tan \left(\arctan \left({e}^{x}\right)\right) = {e}^{x}$

$\implies \frac{d}{\mathrm{dx}} \tan \left(y\right) = \frac{d}{\mathrm{dx}} {e}^{x}$

$\implies {\sec}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / {\sec}^{2} \left(y\right)$

If we draw a right triangle with an angle $y$ and legs such that $\tan \left(y\right) = {e}^{x}$, then we find that $\sec \left(y\right) = \sqrt{{e}^{2 x} + 1}$. Using that, we get our final result:

$\frac{d}{\mathrm{dx}} \arctan \left({e}^{x}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

$= {e}^{x} / {\sec}^{2} \left(y\right)$

$= {e}^{x} / \left({e}^{2 x} + 1\right)$