Given $f(x) = a^x$, one can compute the derivative of $f(x)$ using the chain rule quickly by noticing that
$f(x) = a^x = \left(e^{ln(a)}\right)^x$.
But how would you go about computing the derivative of $f(x)$ without the chain rule, with just the $f'(x) = \lim_\limits{h \to 0} \frac{f(x + h) – f(x)}{h}$ or the $f'(x) = \lim_\limits{x \to a} \frac{f(x) – f(a)}{x – a}$ definition of a derivative and the fact that $\lim_\limits{h \to 0} \frac{e^h – 1}{h} = 1$?
$f'(x) = \lim_\limits{h \to 0} \frac{a^{x + h} – a^x}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{a^h – 1}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{\left(e^{ln(a)}\right)^h – 1}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{(e^h)^{ln(a)} – 1}{h}$.
How would you proceed from here?
Best Answer
Here is a way that uses our knowledge that $\lim_{h\to0}\frac{\mathrm e^h-1}{h}=1$.
At some point in the calculation you demonstrate, we get to the part where we need to evaluate
$$\lim_{h\to0}\frac{\left(\mathrm e^{\ln(a)}\right)^h-1}{h}.$$
Here we can use $(\mathrm{e}^{\ln a})^h=\mathrm e^{h\ln a}$ and substitute $u:=h\ln a$ and note that $u\to0\Leftrightarrow h\to0$. So we are left evaluating
$$\begin{align}\lim_{u\to0}\frac{\mathrm e^u-1}{\frac{u}{\ln a}}&=\lim_{u\to0}\ln(a)\frac{\mathrm e^u-1}{u}\\ &=\ln a\lim_{u\to0}\frac{\mathrm e^u-1}{u}\\ &=\ln (a)\cdot 1\\ &=\ln (a). \end{align}$$