Why aren't the derivatives of these two functions equivalent ($e^{2x}$), ($e^{{x}^2}$)? If $e^{x^{2}}$ is equivalent to $e^{2x}$, does this mean the derivatives of the two functions are equivalent? $2 \cdot e^{2x}$ seems different to $2x \cdot e^{{x}^2}$
 A: You should write $e^{2x}=(e^x)^2$. When you use notation $a^{b^c}$ it means that $b^c$ is the first operation done.
When you write it correctly, you can easily see that the derivatives are the same.
$\frac{d}{dx}e^{2x} = 2e^{2x}$,
$\frac{d}{dx}(e^x)^2 = 2e^x\cdot e^x =2e^{2x}$
A: The issue is that $e^{x^2} \neq e^{2x}$. More precisely, what you can say is $(e^x)^2 = e^{2x}$. We have: $$(e^{x^2})' = 2x e^{x^2}, \quad (e^{2x})' = 2 e^{2x}.$$
A: I'm late to the party, but here is a very important point I don't see addressed very clearly in the other answers:
Why do you think $e^{x^{2}}$ and $e^{2x}$ are equivalent?  Is it because of that rule "when you raise a power to a power, you multiply powers"?
Well, that's only true if you raise a base with a power to another power.
What I mean is:
You can apply the rule to: $(2^{3})^{4}$
But you can't apply the rule to $2^{3^{4}}$.  Why?
Well, $(2^{3})^{4} = 2^{3}2^{3}2^{3}2^{3} = 2^{3 + 3 + 3 + 3} = 2^{3 \cdot 4}$.
But $2^{3^{4}} = 2^{81}$ which is definitely not equal to $2^{12}$.
So, what does this tell you?  It tells you that exponentiation is not associative, that is, we don't have $x^{y^{z}} = (x^{y})^{z}$.
That means you can only apply your rule of "if power raised to another power, multiply the powers" to when you have a base and its power all raised to another power.
We have that with $(e^{x})^{2}$ which the rule tells us equals $e^{2x}$.  But we don't have that with $e^{x^{2}}$, so we can't multiply the powers.  In the latter, we are raising $e$ to the power $x^{2}$, while in the former we raised $e^{x}$ to the power $2$.
A: They are not equivalent.  For $x=1$, note that $e^{1^{2}}=e^{1}=e$ and $e^{2\cdot 1}=e^{2}\neq e$.  Furthermore, their derivatives are not equivalent: $$\frac{d}{dx}e^{x^{2}}=2xe^{x^{2}}$$ and $$\frac{d}{dx}e^{2x}=2e^{2x}$$ and we see that at $x=1$, these have the values $2e$ and $2e^2$, respectively.  
