# How do you differentiate f(x)=e^(tansqrtx) using the chain rule.?

Apr 12, 2016

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

#### Explanation:

Chain rule is basically a lot of substitution.

let $u = \tan \left(\sqrt{x}\right)$

now $f \left(x\right) = {e}^{u}$

so if we try to take the derivative now we apply chain rule to get:
$f ' \left(x\right) = {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

so now we are on a mission to find $\frac{\mathrm{du}}{\mathrm{dx}}$...

We are going to need to rewrite our $u$ equation to something like this:
$u = \tan \left(w\right)$
$w = \sqrt{x}$

Derive $u$

$\frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} w \cdot \frac{\mathrm{dw}}{\mathrm{dx}}$

Now to find $\frac{\mathrm{dw}}{\mathrm{dx}}$...

$w = \sqrt{x}$
Yay we know how to do this one!

$\frac{\mathrm{dw}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$

now substitute back into $\frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} w \cdot \frac{1}{2 \sqrt{x}} = {\sec}^{2} \frac{w}{2 \sqrt{x}}$

and finally substitute back into $f ' \left(x\right)$

$f ' \left(x\right) = {e}^{u} \cdot {\sec}^{2} \frac{w}{2 \sqrt{x}}$

now change everything back to terms of x (you could have done this step earlier)

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