derivative of bases other than e | ap calculus bc,12th grade math
About us/Disclaimer  |  Contact Us

Derivative of bases other than e

Covid-19 has led the world to go through a phenomenal transition .

E-learning is the future today.

Stay Home , Stay Safe and keep learning!!!

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.

Let 'a' be any positive real number and 'u' be a differentiable function of x.
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) $\frac{\text{d}}{\text{d}x}(\log_{a}{u})= \frac{1}{ln(a).u}\frac{\text{d}u}{\text{d}x}$


1) Differentiation of $a^{x}$ (a>0, $a\neq1$) with respect to x is $a^{x}.(ln (a))

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

Examples on Derivative of bases other than e

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

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

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

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

12th grade math

Home

Covid-19 has affected physical interactions between people.

Don't let it affect your learning.

privacy policy