The derivative of exponential function with respect to a variable is equal to the product of the exponential function and natural logarithm of base of the exponential function. The differentiation of exponential function $a^{\displaystyle x}$ with respect to $x$ can be derived in differential calculus by first principle.

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.

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

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

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

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

Use product rule of exponents with same base to split the first term in the numerator of the exponential function.

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

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.

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

Now, factorize the function as two different functions.

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

Use, the product rule of limits for calculating the limit of the function by the product of their limits.

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

Now, evaluate the limit of the exponential function $a^{\displaystyle x}$ as $h$ approaches zero by using direct substitution method.

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

As per the limit of an exponential function, the limit of $\dfrac{a^{\displaystyle h}-1}{h}$ as $h$ approaches zero is equal to natural logarithm of $a$.

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

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

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 natural logarithm of $a$.

Latest Math Topics

Latest Math Problems

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

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

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved