Derivative of Exponential Function

In this tutorial we shall find the derivative of exponential function. We prove the general rules for differentiation of exponential functions.

A function defined by f defined by f\left( x \right) = {a^x},\,\,\,a > 0,\,\,\,a  \ne 1 and x is any real number is called an exponential function.

Let us suppose that the function of the form y = f\left( x \right) =  {a^x}, where a > 0,\,\,a \ne 1

First we take the increment or small change in the function.

\begin{gathered} y + \Delta y = {a^{x + \Delta x}} \\ \Delta y = {a^{x + \Delta x}} - y \\ \end{gathered}

Putting the value of function y = {a^x} in the above equation, we get

\Delta  y = {a^{x + \Delta x}} - {a^x}

Dividing both sides by \Delta  x, we get

\frac{{\Delta  y}}{{\Delta x}} = \frac{{{a^{x + \Delta x}} - {a^x}}}{{\Delta x}}

Taking limit of both sides as \Delta x \to 0, we have

\begin{gathered} \mathop {\lim }\limits_{\Delta x \to 0}  \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0}  \frac{{{a^{x + \Delta x}} - {a^x}}}{{\Delta x}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop  {\lim }\limits_{\Delta x \to 0} \frac{{{a^x}\left( {{a^{\Delta x}} - 1}  \right)}}{{\Delta x}} \\ \end{gathered}

Using the following relation from limit \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} -  1}}{x} = \ln a, we have

\begin{gathered} \frac{{dy}}{{dx}} = {a^x}\ln a \\ \Rightarrow \frac{d}{{dx}}{a^x} = {a^x}\ln a \\ \end{gathered}

Example: Find the derivative of

y = f\left( x \right)  = {4^{2{x^3}}}

We have the given function as

y =  {4^{2{x^3}}}

Differentiation with respect to variable x, we get

\frac{{dy}}{{dx}}  = \frac{d}{{dx}}{4^{2{x^3}}}

Using the rule, \frac{d}{{dx}}{a^x}  = {a^x}\ln a, we get

\begin{gathered} \frac{{dy}}{{dx}} = {4^{2{x^3}}}\ln  4\frac{d}{{dx}}\left( {2{x^3}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} =  {4^{2{x^3}}}\ln 4\left( {6{x^2}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} =  6{x^2}{4^{2{x^3}}}\ln 4 \\ \end{gathered}