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I understand the majority of the proof of the derivative formula for exponential functions of the form: (full proof at bottom of post)

$\frac{d}{dx}a^x$

but I have a little trouble with the last step which implies that the rate of change of any exponential function if proportional to both the function itself and the derivative of the function at zero. I fail to see how

$\lim_{h\to 0}{\frac{a^h-1}{h}}=f'(0)$

*the limit is supposedly the value of the derivative of $a^x$ at $0$, but to me the limit seems only to be a simplified part of the whole definition of the derivative of the function! Where is the flaw in this line of reasoning?

(to clarify my question: It seems like the $lim_{h\to 0}$ defines the derivative of $f$ with respect to $x$ anywhere, not just $0$.)

Proof:

$f(x)=a^x$

$f'(x)=\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}}$

$f'(x)=\lim_{h\to 0}{\frac{a^{x+h} -a^x}{h}}$

$=\lim_{h\to 0}{\frac{a^x(a^h-1)}{h}}$

$=a^x \cdot \lim_{h\to 0}{\frac{a^h-1}{h}}$ , and supposedly $f'(0)=\lim_{h\to 0}{\frac{a^h-1}{h}}$, so

$f'(x)=f'(0)\cdot a^x$

$\square$

I also apologize to anyone who finds this question too basic; I just can't seem to break this mental block. Any help from a deeper understanding is greatly appreciated!

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    $\begingroup$ $$f'(0) = \lim_{h\to 0} \frac{f(0+h) - f(0)}{h}.$$ What is the quotient on the right? $\endgroup$ – Daniel Fischer Apr 29 '16 at 15:59
  • $\begingroup$ You have said $\displaystyle f'(x)=\lim_{h\to 0}{\dfrac{a^{x+h} -a^x}{h}}$, so letting $x=0$ gives $\displaystyle f'(0)=\lim_{h\to 0}{\dfrac{a^{h} -a^0}{h}}$, and $a^0=1$ $\endgroup$ – Henry Apr 29 '16 at 15:59
  • $\begingroup$ "...but to me the limit seems only to be a simplified part of the whole definition of the derivative of the function!" Your sentence says "supposedly A, but B" where both A and B are correct. There is no tension between the two or as to the validity of either one on its own. $\endgroup$ – zyx Apr 29 '16 at 16:09
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The intent of the proof is to relate the derivative at $x$ to the derivative at $0$, which is possible thanks to the property $a^{x+y}=a^xa^y$.

It states

$$f'(x)=(a^x)'=\lim_{h\to0}\frac{a^{x+h}-a^x}h=\lim_{h\to0}a^x\frac{a^{h}-1}h=a^x\lim_{h\to0}\frac{a^{h}-1}h=a^xL_a.$$

And obviously, setting $x=0$, you have $f'(0)=(a^x)'|_{x=0}=a^0L_a=L_a$.

This is a major finding, as you can compute the derivative via a constant,

$$f'(x)=(a^x)'=a^xL_a=a^xf'(0).$$

Of course, this doesn't say what $L_a$ is (not even if it exists), but what matters at this stage is that there is no dependency on $x$.

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By definition, $$f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}.$$ For $a > 0$ and $f(t) = a^t$, we get $$f'(t) = \lim_{h \to 0} \frac{a^{t+h} - a^t}{h}.$$ Now let $t = 0$: $$f'(0) = \lim_{h \to 0} \frac{a^{0+h} - a^0}{h} = \lim_{h \to 0} \frac{a^h - 1}{h}.$$

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The limit is the same as the one that defines $f'(0)$ using the definition of derivative applied to this function $f$. The harder part of the proof, not written in the question, is to show that this limit exists and to determine its value (a logarithm) as a function of the base $a$. The calculation posted in the question is the easier algebraic step of proving $f'(x) = f(x) f'(0)$.

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