# How do you differentiate f(x)=sec(e^(sqrtx-4) )  using the chain rule?

Jan 23, 2016

$f ' \left(x\right) = \frac{{e}^{\sqrt{x} - 4} \sec \left({e}^{\sqrt{x} - 4}\right) \tan \left({e}^{\sqrt{x} - 4}\right)}{2 \sqrt{x}}$

#### Explanation:

The chain rule states that $\frac{d}{\mathrm{dx}} \left[\sec \left(u\right)\right] = u ' \cdot \sec \left(u\right) \tan \left(u\right)$, and here $u = {e}^{\sqrt{x} - 4}$.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[{e}^{\sqrt{x} - 4}\right] \cdot \sec \left({e}^{\sqrt{x} - 4}\right) \tan \left({e}^{\sqrt{x} - 4}\right)$

To find this derivative, use the rule that $\frac{d}{\mathrm{dx}} \left[{e}^{u}\right] = u ' \cdot {e}^{u}$, and this time $u = \sqrt{x} - 4$.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\sqrt{x} - 4\right] \cdot {e}^{\sqrt{x} - 4} \sec \left({e}^{\sqrt{x} - 4}\right) \tan \left({e}^{\sqrt{x} - 4}\right)$

To find $\frac{d}{\mathrm{dx}} \left[\sqrt{x} - 4\right]$, use the power rule, in that $\frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{2}} - 4\right] = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$. This gives

$f ' \left(x\right) = \frac{{e}^{\sqrt{x} - 4} \sec \left({e}^{\sqrt{x} - 4}\right) \tan \left({e}^{\sqrt{x} - 4}\right)}{2 \sqrt{x}}$