Derivative of $f(x)= (\sin x)^{\ln x}$ I am just  wondering if i went ahead to solve this correctly? 
I am trying to find the derivative of $f(x)= (\sin x)^{\ln x}$
Here is what i got below.
$$f(x)= (\sin x)^{\ln x}$$
$$f'(x)=\ln x(\sin x) \Rightarrow f'(x)=\frac{1}{x}\cdot\sin x + \ln x \cdot \cos x$$
Would that be the correct solution?

Since
$$
f(x)=\exp[\ln(x)\ln(\sin x)],
$$it
 follows that
$$
f'(x)=[\ln(x)\ln(\sin x)]'f(x)=\left(\frac{\ln(\sin x)}{x}+\cot(x)\ln x\right)(\sin x)^{\ln x}.
$$
