Proof : Let f(x) = $a^{x}$. Then f(x + h) = $a^{x + h}$
$\frac{\text{d}}{\text{d}x}(f(x))= \lim_{h \rightarrow 0}\frac{f( x+ h) -f(x)}{h}$
$\frac{\text{d}}{\text{d}x}(a^{x})= \lim_{h \rightarrow 0}\frac{a^{x + h} -a^{x}}{h}$
$= \lim_{h \rightarrow 0}\frac{a^{x}.a^{h} -a^{x}}{h}$
$= a^{x}\lim_{h \rightarrow 0}\frac{a^{h} -1}{h}$ [ since $\lim_{h \rightarrow 0}\frac{a^{h} -1}{h} = ln(a) ]$
$\frac{\text{d}}{\text{d}x}(a^{x})= a^{x}.ln (a)$
2) Differentiation of $a^{u}$ with respect to x is $a^{u}. ln(a).\frac{\text{d}u}{\text{d}x}$
Proof : f(x) = $a^{u}$
by using the chain rule
Let u = x
$\frac{\text{d}u}{\text{d}x}$ = 1
Now, $\frac{\text{d}}{\text{d}x}(a^{u}) = \frac{\text{d}}{\text{d}x}(a^{x}) $
=$a^{x}.ln (a)$.1
= $a^{x}.ln (a).\frac{\text{d}u}{\text{d}x}$
Now replace x = u
$\frac{\text{d}}{\text{d}x}(a^{u}) = a^{u}.ln (a).\frac{\text{d}u}{\text{d}x}$
3) Differentiation of $\log_{a}{x}$ ( a>0, a $\neq$ 1) with respect to x is $\frac{1}{ln(a).x}$
Proof : Let f(x) = $\log_{a}{x}$ then
f(x+ h) = $\log_{a}{(x + h)}$
$\frac{\text{d}}{\text{d}x}(f(x))= \lim_{h \rightarrow 0}\frac{f( x+ h) -f(x)}{h}$
$\frac{\text{d}}{\text{d}x}(\log_{a}{x})= \lim_{h \rightarrow 0}\frac{\log_{a}{(x + h)} -\log_{a}{(x)}}{h}$
=$\lim_{h \rightarrow 0}\frac{\log_{a}{(\frac{x +h }{x}})}{h}$
=$\lim_{h \rightarrow 0}\frac{\log_{a}{(1 +h/x})}{h}$
=$\lim_{h \rightarrow 0}\frac{\log_{a}{(1 +h/x})}{1.h}$
=$\lim_{h \rightarrow 0}\frac{\log_{a}{(1 +h/x})}{ln(a).h}$ [ ln(a) = 1]
Divide by x, in short adjust x in such a way that in the denominator we will have h/x
=$\lim_{h \rightarrow 0}\frac{\log_{a}{(1 +h/x})}{x.(h/x)}$
=$\frac{1}{x}\lim_{h \rightarrow 0}\frac{\log_{a}{(1 +h/x})}{(h/x)}$
= $\frac{1}{x}. \frac{1}{ln(a)}$ [ since $\lim_{h \rightarrow 0}\frac{\log_{a}{(1 +h/x})}{(h/x)}$ = 1 ]
$\frac{\text{d}}{\text{d}x}(\log_{a}{x}) = \frac{1}{x.ln(a)}$
Note : Using the chain rule we can prove that : $\frac{\text{d}}{\text{d}x}(\log_{a}{u})= \frac{1}{ln(a).u}\frac{\text{d}u}{\text{d}x}$
Example 1 : $\frac{\text{d}}{\text{d}x}(2^{x})= ln(2).2^{x}$
Example 2 : $\frac{\text{d}}{\text{d}x}(2^{4x})= ln(2).2^{4x}.\frac{\text{d}}{\text{d}x}(4x)$
$\frac{\text{d}}{\text{d}x}(2^{4x})= ln(2).2^{4x}.4 $
$\frac{\text{d}}{\text{d}x}(2^{4x})= 4.ln(2).2^{4x}$
Example 3: Find the derivative of $\frac{\text{d}}{\text{d}x}(\log_{5}{sin(x)})$
Solution : Let u = sin(x)
$\frac{\text{d}u}{\text{d}x}$ cos(x)
By using chain rule
$\frac{\text{d}}{\text{d}x}(\log_{5}{u}) = \frac{1}{u.ln(5)}.\frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}}{\text{d}x}(\log_{5}{sin(x)}) = \frac{1}{sin(x).ln(5)}. cos(x)$
$\frac{\text{d}}{\text{d}x}(\log_{5}{sin(x)}) = \frac{cos(x)}{sin(x).ln(5)}$