How would you go about differentiating this? I tried using the product rule and this is what I have:
but I saw online that the answer is What did I do wrong?
Edit: I have only touched on the basics of differentiation in class, never learned what the Function e is.
Fixed it.
"e" is not a function, it is a number. Euler's number. It has the property that when used as the base of an exponential function, $\displaystyle \begin{align*} e^x \end{align*}$, this function is equal to its derivative. So $\displaystyle \begin{align*} \frac{d}{dx} \left( e^x \right) = e^x \end{align*}$.
So "e" is a number, $\displaystyle \begin{align*} e^x \end{align*}$ is a function.
We start by defining the number e. (e is not a function, but a number.) The simplest definition for our purposes is:
$e\ is\ the\ number\ such\ that\ \displaystyle \lim_{z \rightarrow 0}\dfrac{e^z - 1}{1} = 1.$ Proving that such a number exists is not obvious, but take it on faith.
Now consider the function $y = e^x.$
$e^{(x + h)} = e^x * e^h \implies$
$e^{(x + h)} - e^x= e^x * e^h - e^x = e^x(e^h - 1) \implies$
$\dfrac{e^{(x + h)} - e^x}{h} = \dfrac{e^x(e^h - 1)}{h} = e^x * \dfrac{e^h - 1}{h} \implies$
$\displaystyle \lim_{h \rightarrow 0}\dfrac{e^{(x + h)} - e^x}{h} = \lim_{h \rightarrow 0}\left(e^x * \dfrac{e^h - 1}{h}\right) = \lim_{h \rightarrow 0}\left(e^x\right) * \lim_{h \rightarrow 0}\left(\dfrac{e^h - 1}{h}\right) = e^x * 1 = e^x.$
In other words $j(x) = j'(x) \iff j(x) = e^x.$
Now try differentiating.