Using the definition of the natural logarithm as $\displaystyle\ln x=\int_1^x\frac{dt}{t}$, is there a way to prove that $\frac{d}{dx}a^x=a^x\ln a$?
Proofs that I have generally seen have used the definition of the natural logarithm as the inverse of the exponential function $e^x$: $$\frac{d}{dx}a^x=\frac{d}{dx}e^{x\ln a}=e^{x\ln a}\ln a=a^x\ln a$$ using the chain rule and the fact that $f(x)=e^x$ satisfies $f'(x)=f(x)$. But is there a way to go about finding this derivative without using the definition of the natural logarithm as the inverse of the exponential function?
Would you like to see something like this?
If we define $\exp: \mathbb{R} \to \mathbb{R}_{+}$ as the inverse of the $\log$ function, we get that, by the derivative of the inverse function and the FTC1: $$\exp'(x)=\frac{1}{\frac{1}{\exp(x)}}=\exp(x)$$ And with the following definition of $a^x$ (for $a \neq 1$ and $a > 0$): $$a^x:=\exp(\log(a)x)$$ and the chain rule we get the result you mentioned.