Find $\frac{d}{dx}(e^{\sin(4x)})$ I have the answer to the question and I have the idea of how to get there but I am not connecting all the dots.
$$\frac{d}{dx}e^{\sin(4x)}$$
We have a few forms to pick from at this point, one for natural logs, one for trig functions, cos and sin, and of course exponents.  The exponent formula is $\frac{d}{dx}\left(e^{f(x)}\right)=e^{f(x)}f'(x)$ Then, I start to answer the right hand side of the equation. $$f(x)=\sin(4x)$$ $$f'(x)= \cos(4x)$$
$f'(x)= \cos(x)$ because $\frac{d}{dx}\sin(x)=\cos(x)$.  Now that I have the right hand side entered I write it down. $$\frac{d}{dx}e^{\sin(4x)}=e^{\sin(4x)}\left(\cos(4x)\right)$$  I still have not found the derivative.  There is a cos(f(x)) that is floating around.  What is the rule now?  It can't be the standard trig rule of $\frac{d}{dx}\cos(x)=-\sin(x)$.  Instead there is some other rule.  I know if I take the derivative of $4x$, the answer is $4$. This can be done by $$\frac{d}{dx}4x=\frac{4(x+h)-4x}{h}$$   The answer then after removing like terms and simplifying is $$\frac{d}{dx}4x=4$$
That would be helpful because the answer to the problem is $$\frac{d}{dx}e^{\sin(4x)}=4e^{\sin(4x)}\left(\cos(4x)\right)$$  I feel like I am close but I am not really understanding why I choose to not to convert back to $- \sin(x)$ from $\cos(x)$ and instead only solve for $4x$.  That is where I am lost.
 A: Let’s walk through the chain rule step by step.
Recall that the chain rule states that $$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$$
For your derivative we have $$ \frac{d}{dx} (e^{\sin(4x)})=\frac{d}{d(\sin(4x))}( e^{\sin(4x)})\cdot\frac{d}{dx}(\sin(4x))$$
The first term just returns itself times the natural log of e, which is 1. For the second term, we apply the chain rule again to get
$$ \frac{d}{dx}( e^{\sin(4x)})=e^{\sin(4x)}\cdot\frac{d}{d(4x)}(\sin(4x))\cdot \frac d{dx}(4x)=e^{\sin(4x)}\cdot\cos(4x)\cdot4$$
A: Noticed that you have the composition $(f\circ g\circ h)(x)$ of three functions, then the chain rule says that for derivable functions we have
$$\frac{d(f\circ g\circ h)}{dx}(x)=\frac{df}{dx}(g(h(x)))\cdot \frac{dg}{dx}(h(x))\cdot \frac{dh}{dx}(x),\quad (*)$$
That is just the idea of "chain" rule.
Taking $f(x)=e^{x}, g(x)=\sin(x)$ and $h(x)=4x$ we have $$(f\circ g\circ h)(x)=f(g(h(x)))=e^{g(h(x))}=e^{\sin(h(x))}=e^{\sin(4x)},$$
and using the chain rule $(*)$ we have
$$\frac{d}{dx}e^{\sin(4x)}=\frac{d(f\circ g\circ h)}{dx}(x)=e^{\sin(4x)}\cdot \cos(4x)\cdot 4.$$
Therefore,
$$\frac{d}{dx}e^{\sin(4x)}=4\cos(4x)e^{\sin(4x)}.$$
NB:

*

*Remember that or to show that
$$\frac{d}{dx}e^{x}=e^{x},\quad \frac{d}{dx}\sin(x)=\cos(x),\quad \frac{d}{dx}(4x)=4.$$

*Additionally, for this type of problems in order to seek clarity with the notation. I suggest softening the notation for the chain rule as follows.
$$(f(g(h)))'=f'(g(h))\cdot g'(h)\cdot h'.$$
A: Another way to approach it:
\begin{align*}
y = e^{\sin(4x)} & \Rightarrow \ln(y) = \sin(4x)\\\\
& \Rightarrow \frac{y'}{y} = 4\cos(4x)\\\\
& \Rightarrow y' = 4\cos(4x)y\\\\
& \Rightarrow y' = 4\cos(4x)e^{\sin(4x)}
\end{align*}
Hopefully this helps!
