Differentiate $f(x) = exp_{a}(x) $ from first principles

Question

Differentiate $f(x) = exp_{a}(x) $ from first principles, for $ a > 0 $ (Recall that $ exp_{a}(x) = exp(x.ln(a)) $

Here is where I am so far:

$ f'(x) = \lim\limits_{h \rightarrow 0} \frac{exp((x+h) \cdot ln(a)) - exp(x \cdot ln(a))}{h}$

$ = \lim\limits_{h \rightarrow 0} \frac{exp(x \cdot ln(a)) \cdot exp(h \cdot ln(a)) - exp(c \cdot ln(a))}{h} $

$ = exp(x \cdot ln(a)) \cdot \lim\limits_{h \rightarrow 0} \frac{exp(h \cdot ln(a))-1}{h} $

$ = a^{x} \cdot \lim\limits_{h \rightarrow 0} \frac{a^{h \cdot ln(a)} - 1}{h} $

Answer 1

You have

$$\begin{align} \exp(x\ln(a)) \lim_{h \to 0} \frac{\exp(h\ln(a)) - 1}{h} &= a^{x} \lim_{h \to 0} \frac{\exp(h\ln(a)) - 1}{h} \end{align}$$

Set

$$y = h\ln(a) \implies h = \frac{y}{\ln(a)}$$

and notice that $y \to 0$ as $h \to 0$

Hence

$$\begin{align} a^{x} \lim_{h \to 0} \frac{\exp(h\ln(a)) - 1}{h} &= a^{x} \lim_{y \to 0} \frac{\exp(y) - 1}{\frac{y}{\ln(a)}} \\ &= \ln(a)a^{x} \lim_{y \to 0} \frac{\exp(y) - 1}{y} \ \ \ (*) \\ \end{align}$$

This last limit can be evaluated using L'Hopitals rule as the limit gives $\frac{0}{0}$

$$\begin{align} \implies \lim_{y \to 0} \frac{\exp(y) - 1}{y} &= \lim_{y \to 0} \exp(y) \\ &= 1 \ \ \ \ \ \ (**)\\ \end{align}$$

Substituting $(**)$ into $(*)$ we get

$$f'(x) = \ln(a) a^{x}$$

Answer 2

$f(x)=exp(x\cdot ln(a))=exp(u)$ where $u=x\cdot ln(a)$

Using the chain rule, we get

$f'(x)=exp(u)\cdot u'=exp_a(x)\cdot ln(a)$

Edit: We can use this, but eventually, you're going to have to use the definitions and properties.

$f(x)=exp(x\cdot ln(a))$

$f'(x)=\lim_{h\to 0}\frac{exp((x+h)\cdot ln(a))-exp(xln(a))}{h}$

$=\lim_{h\to0}\frac{exp(xln(a))\cdot exp(hln(a))-exp(xln(a))}{h}$

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

$=a^x\cdot\lim_{h\to0}\frac{a^h-1}{h}$

Let $h=log_a{t}$

$=a^x\cdot\lim_{t\to1}\frac{t-1}{log_a(t)}$

$=a^x\cdot\lim_{t\to1}\frac{t-1}{ln(t)-ln(1)}\cdot \frac{ln(t)}{\log_a{t}}$

$=a^x\cdot(ln(t))'|_{t=1}\cdot ln(a)$

$=ln(a)a^x$