First derivative of multiplied powers Wolfram Alfa shows $\frac{d}{dx}e^{4y} = 4e^{4y}$
but I do not understand how to get to that answer
I have
$e^{4y} = (e^4)^y$
So by the chain rule is it not the case that
\begin{align}
\frac{d}{dx}e^{4y} & = y(e^4)^{y - 1}\cdot4e^3 \\
& = y \cdot e^{4y - 4} \cdot 4e^3 \\
& = 4ye^{4y-1} \\
& = \frac{4ye^{4y}}{e}
\end{align}
Clearly somewhere I am making a big mess
 A: $\frac d{dx}e^x=e^x$
$\frac d{dx}e^y=y'e^y$ - which is an application of the chain rule
$\frac d{dx}a^x=\frac d{dx}e^{x\ln a}=\ln a \cdot e^{x\ln a}=\ln a \cdot a^x$
$\frac d{dx} (e^4)^x=\ln(e^4)\cdot e^{x\ln (e^4)}=4e^{4x}$
A: The first thing to note is that you have the derivative with respect to $x$, but $x$ is not in your equation. Technically: 
$\frac{d}{dx}e^{4y} = 0$
You should write: 
$\frac{d}{dy}e^{4y}$
Then we can get that with the chain rule. Let's go through the steps explicitly.
$ F = e^{4y}$
to:
$ F = u^{y} $
by saying that $u = e^{4}$.
The chain rule would then be written as:
$ \frac{dF}{dy} = \frac{dF}{du}\frac{du}{dy} $
You have the right result for $\frac{dF}{du}$, but you have an error in $\frac{du}{dy}$. The function $u$ does not contain the variable $y$, so its derivative is zero. That gives you the answer of 0 overall. Clearly that's not right.
What you need to do, when applying the chain rule, is pick a $u$ that is a function of $y$. If you pick:
$u = 4y$, then you get:
$F = e^{u}$
Using the exact same chain rule formula, you can get $\frac{dF}{du}$ and then $\frac{du}{dy}$
