I doubt this is a suitable problem for a novice. even showing convergence is tough. i will look at the other posted answers. there is a good reason people start from the integral definition of ln(x) to derive this result.
If f(x)= a^{x}, the f(x+h)= a^{a+x}= a^{x}a^{h} so f(a+ h)- f(a)= a^{x}(a^{h}- 1). The derivative is lim (f(x+h)- f(x))/h= a^{x}lim {(a^{h}-1)/h}. Notice that that is a^{x} time a limit that is independent of x. That is, as long as the derivative exists, it is a^{x} times a constant. The problem is showing that the lim{(a^{h}-1)/h} EXISTS! And then showing that, if a= 2, that limit is ln(2). Showing that that limit exists is sufficiently non-trivial that many people (myself included), as mathwonk said, prefer to define ln(x) as the integral, from 1 to x of (1/t)dt. From that, it is possible to prove all properties of ln(x) including (trivially) that the derivative is 1/x. Defining e^{x} as the inverse function of ln(x) leads to all the properties of e^{x} (including the fact that it is some number to a power!), in particular that its derivative is e^{x} itself and, from that, that the derivative of a^{x} is (ln a) a^{x}.
No, you did exactly what HallsofIvy was advocating, he was just pointing out that the question asked for it to be solved using the definition of a derivative, which makes things much harder. Easier to approach things from the other way, starting by defining the integral of 1/x.
"The derivative is lim (f(x+h)- f(x))/h= axlim {(ah-1)/h}. Notice that that is ax time a limit that is independent of x. That is, as long as the derivative exists, it is ax times a constant. The problem is showing that the lim{(ah-1)/h} EXISTS! And then showing that, if a= 2, that limit is ln(2)." tell me if i'm wrong, but it doesn't seems so hard to determine this limit.. (a^h-1)/h = (exp (h*ln(a) )-1) / h = ( 1 + h*ln(a) + o(h*ln(a)) - 1 ) / h h->0 = ln(a) + o(ln(a)) so lim (a^h-1)/h = ln(a) .......
Yes, assuming that you know "(exp (h*ln(a) )-1) / h= ( 1 + h*ln(a) + o(h*ln(a)) - 1 ) / h " its easy to do it. Proving what you assumed is the hard part!