Q:

Uniqueness of exponential function

To my knowledge, the exponential function is the unique function satisfying
$f'=f$ and
$f(0)=1$
however, unless I've made a mistake, we have
$$\frac{\partial}{\partial x} (ax)^x = x (ax)^{x-1} a = ax (ax)^{x-1} = (ax)^x$$
and 
$$(a0)^0 = 0^0 =1$$
so I feel like I must be missing something special about $e^x$.  Any pointers would be greatly appreciated.

A:

You differentiated $x^x$ wrong. In fact,
$$  (x^x)' = (e^{x \log x})' \overset{\text{chain rule}}{=} [x \log x]' e^{x\log x} = (\log x +1 )x^x$$
The rule $[x^n]' = n x^{n-1}$ only applies when $n$ is a fixed constant.

A:

You've made a mistake. Distribute $x $
$(ax)^x = a^x x^x $
Now can what you've written for the derivative be true?

A:

The rule that $\dfrac d{dx} x^n = nx^{n-1}$ holds when $n$ is constant, i.e. $n$ does not change as $x$ changes.  In the case of $(ax)^x$, the exponent changes as $x$ changes, and the power rule is not applicable.  You can use logarithmic differentiation in such a case.

