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?

Five different answers, and all of them using exponentials and logarithms? While logs are indeed convenient, it's certainly possible to solve this problem without what one might consider a detour into logarithm-land.
We know that $(x^n)'=nx^{n-1}$ and $(a^x)'=a^x\log a$, which means that
$$(y^z)'=zy^{z-1}y'$$
when $z$ is constant, and
$$(y^z)'=y^{z}\log y\cdot z'$$
when $y$ is constant. Therefore, when both $y$ and $z$ vary, we have
$$(y^z)'=zy^{z-1}y'+y^{z}\log y\cdot z'.$$
Substitute $y = \sin x$ and $z = \log x$, and you're done.
