Trouble with Logarithmic Differentiation

Hey guys I'm trying to find the derivative of this equation using logarithmic differentiation but I'm having some trouble. Wolfram Alpha is giving me different answers and I'm having difficulty understanding the steps it spits out. Could somebody take a look at what I'm doing and tell me where I went wrong?

The problem I need to differentiate is:

$y=(\mathrm{sin}x{)}^{\mathrm{ln}x}$

My first step is to take the logarithm of both sides and factor out the exponent.

$\mathrm{ln}(y)=\mathrm{ln}x\ast \mathrm{ln}(\mathrm{sin}x)$

Next I try to differentiate both sides, using the rules for logarithms:

$\frac{1}{y}\ast {y}^{\prime}=\frac{1}{x}\frac{1}{\mathrm{sin}x}\mathrm{cos}x\Rightarrow \frac{1}{y}\ast {y}^{\prime}=\frac{\mathrm{cos}x}{x\mathrm{sin}x}$

Which finally renders:

$\frac{dy}{dx}=\frac{\mathrm{cos}x}{x\mathrm{sin}x}(\mathrm{sin}x{)}^{\mathrm{ln}x}$

But this doesn't seem to be the right answer, could somebody help me out?

Hey guys I'm trying to find the derivative of this equation using logarithmic differentiation but I'm having some trouble. Wolfram Alpha is giving me different answers and I'm having difficulty understanding the steps it spits out. Could somebody take a look at what I'm doing and tell me where I went wrong?

The problem I need to differentiate is:

$y=(\mathrm{sin}x{)}^{\mathrm{ln}x}$

My first step is to take the logarithm of both sides and factor out the exponent.

$\mathrm{ln}(y)=\mathrm{ln}x\ast \mathrm{ln}(\mathrm{sin}x)$

Next I try to differentiate both sides, using the rules for logarithms:

$\frac{1}{y}\ast {y}^{\prime}=\frac{1}{x}\frac{1}{\mathrm{sin}x}\mathrm{cos}x\Rightarrow \frac{1}{y}\ast {y}^{\prime}=\frac{\mathrm{cos}x}{x\mathrm{sin}x}$

Which finally renders:

$\frac{dy}{dx}=\frac{\mathrm{cos}x}{x\mathrm{sin}x}(\mathrm{sin}x{)}^{\mathrm{ln}x}$

But this doesn't seem to be the right answer, could somebody help me out?