# What is the derivative of this function (cos(e^(x^5*sin(x))))^(1/2)?

May 28, 2018

(-x^4e^(x^5sinx)sin(e^(x^5sinx))(xcosx + 5sinx))/(2sqrt(cos(e^(x^5sinx)))

#### Explanation:

Given: $y = {\left(\cos \left({e}^{{x}^{5} \sin x}\right)\right)}^{\frac{1}{2}}$

Use the chain rule to differentiate.

Derivative rules needed:

Power rule: $\text{ } \left({u}^{n}\right) ' = n {u}^{n - 1}$

(cos u)' = -u' sin u; " "(e^u)' = u' e^u

product rule: $\text{ } \left(u v\right) ' = u v ' + v u '$

Start out with the power rule where u = cos(e^(x^5sinx)); n = 1/2:

$y ' = \frac{1}{2} {\left(\cos \left({e}^{{x}^{5} \sin x}\right)\right)}^{- \frac{1}{2}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} \left(\cos u\right)$

Let $u = {e}^{{x}^{5} \sin x}$:

$y ' = \frac{1}{2} {\left(\cos \left({e}^{{x}^{5} \sin x}\right)\right)}^{- \frac{1}{2}} \left(- \sin \left({e}^{{x}^{5} \sin x}\right)\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}} \left({e}^{u}\right)$

Let $u = {x}^{5} \sin x$:

$y ' = \frac{1}{2} {\left(\cos \left({e}^{{x}^{5} \sin x}\right)\right)}^{- \frac{1}{2}} \left(- \sin \left({e}^{{x}^{5} \sin x}\right)\right) {e}^{{x}^{5} \sin x} \cdot \frac{d}{\mathrm{dx}} \left({x}^{5} \sin x\right)$

Use the product rule, $u = {x}^{5} , u ' = 5 {x}^{4} , v = \sin x , v ' = \cos x$:

$y ' = \frac{1}{2} {\left(\cos \left({e}^{{x}^{5} \sin x}\right)\right)}^{- \frac{1}{2}} \left(- \sin \left({e}^{{x}^{5} \sin x}\right)\right) {e}^{{x}^{5} \sin x} \left({x}^{5} \cos x + 5 {x}^{4} \sin x\right)$

Rearrange and factor (GCF = ${x}^{4}$):

y' = (-x^4e^(x^5sinx)sin(e^(x^5sinx))(xcosx + 5sinx))/(2sqrt(cos(e^(x^5sinx)))