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

We have $f \left(x\right) = \sin {x}^{\cos} x$ we can write this using logarithms as

$f \left(x\right) = {e}^{\cos x \cdot \ln \sin x}$

hence its first derivative is

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = {e}^{\cos x \cdot \ln \sin x} \cdot \left[\left(- \sin x\right) \cdot \ln \sin x + \cos x \cdot \frac{\cos x}{\sin x}\right]$

or

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \sin {x}^{\cos} x \cdot \left[{\cos}^{2} \frac{x}{\sin} x - \sin x \cdot \ln \sin x\right]$