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Derivative of bases other than e : To differentiate exponential and logarithmic functions to other bases, you have three options:(1) use the definitions of $a^{x} and \log_{a}{x}$ and differentiate using the rules for the natural exponential and logarithmic functions.

(2) use logarithmic differentiation, or

(3) use the differentiation rules for bases other than given in the theorem below.

1) $\frac{\text{d}}{\text{d}x}(a^{x}) = a^{x}. ln(a)$

2) $\frac{\text{d}}{\text{d}x}(a^{u}) = a^{u}. ln(a).\frac{\text{d}u}{\text{d}x}$

2) $\frac{\text{d}}{\text{d}x}(\log_{a}{x})= \frac{1}{ln(a).x}$

3) $\frac{\text{d}}{\text{d}x}(\log_{a}{u})= \frac{1}{ln(a).u}\frac{\text{d}u}{\text{d}x}$

**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

$\frac{\text{d}}{\text{d}x}(\log_{5}{u}) = \frac{1}{u.ln(5)}.\frac{\text{d}u}{\text{d}x}$

$\frac{\text{d}}{\text{d}x}(\log_{5}{sin(x)}) = \frac{1}{sin(x).ln(5)}. cos(x)$

$\frac{\text{d}}{\text{d}x}(\log_{5}{sin(x)}) = \frac{cos(x)}{sin(x).ln(5)}$

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