What you have written is best expressed using notation:
$y=\exp(2\log(sin(x))$, that is using web-convenient notation: $e^t=\exp(t)~\&~ln(x)=\log(x)$
Because $\exp(\log(T))=T$ what you posted is $y=\exp(2\log(\sin(x)))=\exp(\log(\sin^2(x)))=\sin ^2(x)$.
Thus if $y=\sin^2(x)$ then $y'=2(\sin(x))(\cos(x))$
Can you finish ?
Ahh I see thank you. You would just use the identity to rewrite 2sin(x)cos(x) = sin(2x). Could you take a moment to explain why we cannot derive it as we would normally with exp functions. e.g. we would derive y = e^3x as dy/dx = e^3x . 3
Thanks a bunch again
In your attempt you stopped at $\displaystyle \frac{dy}{dx}= 2cot(x)e^{2 ln(sin(x))}= \frac{2cos(x)}{sin(x)}e^{2 ln(sin(x))}$.
Now, as Plato suggested, use $\displaystyle e^{2 ln(sin(x))}= e^{ln(sin^2(x))}= sin^2(x)$ to reduce that to $\displaystyle \frac{dy}{dx}= \frac{2 cos(x)}{sin(x)}sin^2(x)= 2 sin(x)cos(x)$.