Math Doubts

Proof of Derivative of Exponential function by the Limits

The derivative of an exponential function $x$-th power of $a$ with respect to $x$ can be proved by the fundamental definition of the derivatives.

$\dfrac{d}{dx}{\big(a^{\displaystyle x}\big)}$ $\,=\,$ $a^{\displaystyle x} \times \log_{e}{a}$

Let us learn how to derivative the differentiation of the exponential function by the first principle of the derivatives.

Derivative of a function in Limit form

To find the derivative of exponential function $a^{\displaystyle x}$ with respect to $x$, write the derivative of this function in limit form by the definition of the derivative.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f(x+\Delta x)-f(x)}{\Delta x}}$

It is simply written in the following form for our convenience.

$\implies$ $\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

Suppose $f{(x)} \,=\, a^{\displaystyle x}$, then $f{(x+h)} \,=\, a^{\displaystyle x+h}$. Now, substitute them in the first principle of the derivative for evaluating the differentiation of the exponential function $a^{\displaystyle x}$ with respect to $x$ in the differential calculus.

$\implies$ $\dfrac{d}{dx}{\, (a^{\displaystyle x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x+h}-a^{\displaystyle x}}{h}}$

Simplify the exponential function

It is time to simplify the whole function. Look at the exponential expression in the numerator. It consists of two terms. In both terms, there is a common factor. Firstly, use product rule of exponents with same base to split the first term as a product of two exponential functions.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x} \times a^{\displaystyle h}-a^{\displaystyle x}}{h}}$

In numerator, $a^{\displaystyle x}$ is a common factor in the both terms of the expression. So, it can be taken out common from them for simplifying this exponential function mathematically in the numerator.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x}\big(a^{\displaystyle h}-1\big)}{h}}$

The mathematical function can be split as a product of two functions by factorization (or factorisation).

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \bigg(a^{\displaystyle x} \times \dfrac{a^{\displaystyle h}-1}{h}\bigg)}$

Evaluate the Limit of the function

The limit of the function should have to be evaluated when the value of $h$ approaches zero. So, the exponential function $a$ raised to the $x$-th power is a constant. So, it can be taken out from the limit operation as per the constant multiple rule of limits.

$=\,\,\,$ $a^{\displaystyle x}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle h}-1}{h}}$

According to the limit rule of exponential function, the limit of $a$ raised to the power of $h$ minus $1$ by $h$ as the value $h$ is closer to zero is equal to natural logarithm of $a$.

$=\,\,\,$ $a^{\displaystyle x}$ $\times$ $\log_{e}{a}$

$\therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\,\big(a^{\displaystyle x}\big)} \,=\, a^{\displaystyle x}\log_{e}{a}$

Therefore, it is proved that the derivative of exponential function $a^{\displaystyle x}$ with respect to $x$ is equal to product of the exponential function $a^{\displaystyle x}$ and the natural logarithm of $a$.

Math Doubts

A best free mathematics education website that helps students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

A math help place with list of solved problems with answers and worksheets on every concept for your practice.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved