how do you find the derivative of (ln x)^sin x
$\displaystyle y = (\ln{x})^{\sin{x}}$
$\displaystyle \ln{y} = \sin{x} \ln(\ln{x})$
$\displaystyle \frac{y'}{y} = \sin{x} \cdot \frac{\frac{1}{x}}{\ln{x}} + \ln(\ln{x}) \cos{x}$
$\displaystyle \frac{y'}{y} = \frac{\sin{x}}{x \ln{x}} + \ln(\ln{x}) \cos{x}$
$\displaystyle y' = (\ln{x})^{\sin{x}}\left[\frac{\sin{x}}{x \ln{x}} + \ln(\ln{x}) \cos{x}\right]$