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 Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Math Worksheets

The math worksheets with answers for your practice with examples.

Practice now

Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.

Learn more

Math Doubts

A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects.

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved