# How do you find the derivative of y = x^(cos x)?

Feb 18, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\cos} x \left(\frac{\cos x}{x} - \ln x \sin x\right)$

#### Explanation:

Take the natural log of both sides and move the cos in front of the natural log.

$\ln y = \left(\cos x\right) \ln x$

Use implicit differentiation. You have to take the product rule of the right side.

$\left(\frac{1}{y}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \left(\cos x\right) \left(\frac{1}{x}\right) + \left(\ln x\right) \left(- \sin x\right)$

$\left(\frac{1}{y}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{\cos x}{x} - \ln x \sin x$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{\cos x}{x} - \ln x \sin x\right)$

Plug in ${x}^{\cos} x$ for y

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\cos} x \left(\frac{\cos x}{x} - \ln x \sin x\right)$