# What is the derivative of y = (sin x)^(cos x)?

Jan 4, 2017

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

#### Explanation:

Take the natural logarithm of both sides.

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

Use the rule $\log {a}^{n} = n \log a$ to simplify:

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

Use the implicit differentiation as well as the product and chain rules to differentiate.

$\frac{d}{\mathrm{dx}} \left(\ln \sin x\right) = \frac{1}{\sin} x \cdot \cos x = \cos \frac{x}{\sin} x = \cot x$

Now to the relation as a whole:

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

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

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

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

Hopefully this helps!