How do you find the derivative of y=x^cosx?

Aug 20, 2016

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

Explanation:

Take the natural logarithm of both sides:

$\ln y = \ln \left({x}^{\cos x}\right)$

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

Now, take the derivative of both sides.

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

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

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

Hopefully this helps!