How do you prove the derivative of f(x) = a^x?
Attempt:

I got to this step:
(a^x) * (lim h - 0 of (a^h - 1/h))

I realized that if I were to sub in f'(0), I would get a^h - h, so,

= a^x * f'(0)

Now, how do I get from the above to:
f'(x) = a^x*ln(a) ?

Question

How do you prove the derivative of f(x) = a^x?
Attempt:

I got to this step:
(a^x) * (lim h - > 0 of (a^h - 1/h))

I realized that if I were to sub in f'(0), I would get a^h - h, so,

= a^x * f'(0)

Now, how do I get from the above to:
f'(x) = a^x*ln(a) ?

anonymous · Answer

proof using limits..

anonymous · Answer

Yea thats what I started with and then I got this point
= a^(x) * f'(0).

anonymous · Answer

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

anonymous · Answer

Ok I am stuck.

anonymous · Answer

What do I do from where I left off?

anonymous · Answer

$$\lim_{h \rightarrow 0}\frac{a^{x+h}-a^{x}}{h}=a^{x}\lim_{h \rightarrow 0}\frac{a^{h}-1}{h}$$So now we have to show that the limit is equal to ln(a).

anonymous · Answer

How?

anonymous · Answer

Good question.  I'm stuck, too.

anonymous · Answer

The limit as it's written goes to 0/0.  I was thinking about using l'Hopital's rule and taking the derivative of top and bottom wrt h, but then I realized that involved using $$\frac{d}{dh}a^{h}$$We would be here forever.  I think there must be some substitution we could use, but I don't see what it is.