y= e^sin 2x
Solution attempt:
lny = sin 2x
lny = 2sinxcosx
$\displaystyle \dfrac{d}{dx}\left[e^u\right] = e^u \times \dfrac{du}{dx}$
For $\displaystyle y = e^{\sin(2x)}$, let $\displaystyle u = \sin(2x)$. Then, $\displaystyle \dfrac{du}{dx} = 2\cos(2x)$.
Now, it is only a matter of substituting the values of $\displaystyle u$ and $\displaystyle \dfrac{du}{dx}$ into the first equation at the top.
$\displaystyle y' = e^{\sin(2x)} \times 2\cos(2x) = 2e^{\sin(2x)}\cos(2x)$
$\displaystyle y= e^{sin 2x}$
$\displaystyle \ln(y)=\sin(2x)$
$\displaystyle \frac{dy}{ydx}=\frac{d}{dx}\left [ \sin(2x) \right ]$ implicite differentiation.
$\displaystyle \frac{dy}{ydx}=2\cos(2x)$chain rule
$\displaystyle \frac{dy}{dx}=2y\cos(2x)$
$\displaystyle y=e^{sin 2x}$
$\displaystyle \frac{dy}{dx}=2e^{sin 2x}\cos(2x)$
Then, $\displaystyle \dfrac{du}{dx} = 2\cos(2x)$.
Both of you have the correct answer, I get the use of the chain rule and it looks very clean but could you also use the product rule for $\displaystyle \dfrac{du}{dx}$?
Thus making $\displaystyle \dfrac{du}{dx}=2\sin+2x\cos$
did I just make a mistake?
Product rule applies when you have the form: $\displaystyle f(x)=p(x)q(x)$. (which you don't have here)
I am not sure where you are geting 2sin(2x)+2cos(2x) but if you will notice, the form of f(x) needing to use the chain rule is: $\displaystyle f(x)=p(q(x))$
where p(x)=sin(x) and q(x)=2x