First derivative of multiplied powers

Question

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

Answer 1

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 $\frac{du}{dy}$

Answer 2

$\frac d{dx}e^x=e^x$

$\frac d{dx}e^y=y'e^y$ - which is an application of the chain rule

$\frac d{dx}a^x=\frac d{dx}e^{x\ln a}=\ln a \cdot e^{x\ln a}=\ln a \cdot a^x$

$\frac d{dx} (e^4)^x=\ln(e^4)\cdot e^{x\ln (e^4)}=4e^{4x}$