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

Dec 11, 2015

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

#### Explanation:

According to the chain rule,

$f ' \left(x\right) = {e}^{\sec \sqrt{x}} \frac{d}{\mathrm{dx}} \left[\sec \sqrt{x}\right]$

First, find the internal derivative (again using the chain rule).

$\frac{d}{\mathrm{dx}} \left[\sec \sqrt{x}\right] = \sec \sqrt{x} \tan \sqrt{x} \frac{d}{\mathrm{dx}} \left[\sqrt{x}\right]$

$= \frac{1}{2} {x}^{- \frac{1}{2}} \sec \sqrt{x} \tan \sqrt{x}$

$= \frac{\sec \sqrt{x} \tan \sqrt{x}}{2 \sqrt{x}}$

Plug this back in to find $f ' \left(x\right)$.

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

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