Using the chain rule to find derivative Could you help me figure this one out?
$$f(x) = 6e^{x\sin x}$$
$$f'(x) = \,?$$
 A: Add logarithms to both sides of the equation $f(x) = 6e^{x \sin x}$ 
\begin{align}\ln f(x) &= \ln (6e^{x \sin x}) \\
&= \ln 6 + \ln e^{x \sin x} \\
&= \ln 6 + x \sin x,
\end{align}
and implicitly differentiate both sides to get
\begin{align}
\frac{f'(x)}{f(x)}=\sin x + x \cos x.
\end{align}
Finally, we get
\begin{align}
f'(x)&=f(x)(\sin x + x \cos x) \\ &= \boxed{6e^{x \sin x}(\sin x + x \cos x)}.
\end{align}   
A: Take things step-by-step:
Chain rule applies when you have a function of the form $f(x) = g(h(x))$.  So what is $g(x)$ in this case?  What is $h(x)$?
From there, we know $f'(x) = g'(h(x)) \cdot h'(x)$.  In computing $h'(x)$, you may need to apply the product rule, so be cautious.
A: Hint: In the context of the exponential function, the chain rule says
$$\left(e^{\square}\right)' =e^{\square}\cdot \square'$$
Now fill in the box with what is in your problem.
Also, don't forget the multiplier of $6$ out front, but that's easy. That just tags along as a multiplier in the end.
A: A more direct way, without the logarithms approach in my other posted answer:
\begin{align}\frac d{dx}(6e^{x \sin x})&= 6 \frac d{dx}e^{x\sin x} \\
&=6e^{x \sin x} \frac d{dx}(x \sin x) & \text{chain rule} \\
&=\boxed{6e^{x \sin x}(\sin x + x \cos x)} & \text{product rule}\end{align}
A: $(x\sin x)'=\sin x+x\cos x$, so $(6e^{x\sin x})'=6(\sin x+x\cos x)e^{x\sin x}$.
