# How do you differentiate y=e^(xsecx)+e^5?

Oct 22, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[1 + x \tan x\right] \sec x {e}^{x \sec x}$

#### Explanation:

$y = {e}^{x \sec x} + {e}^{5}$

we want

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({e}^{x \sec x}\right) + \frac{d}{\mathrm{dx}} \left({e}^{5}\right)$

now

$\frac{d}{\mathrm{dx}} \left({e}^{5}\right) = 0$

since$\text{ } {e}^{5}$ is a constant

so that leaves us with

$\frac{d}{\mathrm{dx}} \left({e}^{x \sec x}\right)$

now by the chain rule we have the result

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

but we also have he product rule to use

$\frac{d}{\mathrm{dx}} \left(u v\right) = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

so putting all this together

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({e}^{x \sec x}\right) + \frac{d}{\mathrm{dx}} \left({e}^{5}\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \left[\sec x \frac{d}{\mathrm{dx}} \left(x\right) + x \frac{d}{\mathrm{dx}} \left(\sec x\right)\right] {e}^{x \sec x} + 0$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left[\sec x + x \sec x \tan x\right] {e}^{x \sec x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[1 + x \tan x\right] \sec x {e}^{x \sec x}$