# Derivatives of Exponential Functions

Sub Topics
The Exponential function was developed by Napier while researching logarithims. He assumed the base of the logarithm to be e and called it as an exponential number. This number has a constant value given by 2.718. This number is extensively used in Nuclear physics and in the power transmission theory.

## Derivative of ex

The derivative of exis the easiest rule in calculus, $\frac{d}{dx}$ ($e^{x}$ ) = $e^{x}$

Proof:

Let f( x ) = ex. then f ( x + h ) = $e^{x + h}$, we know

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{[f(x + h) – f(x)]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$  $\frac{[e^{x+h} – e^{x}]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$  $\frac{[e^{x}*e^{h} – e^{x}]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{e^{x}*[e^{h}–1]}{h}$

$\frac{d}{dx}$ f( x ) = $e^{x}$*$lim_{h\rightarrow 0}$ $\frac{[e^{h} – 1]}{h}$

$\frac{d}{dx}$ f( x ) = $e^{x}$ * ( 1 )

$\frac{d}{dx}$ f( x ) = $e^{x}$

## e2x

Find the derivative of $e^{2x}$ with respect to ‘ x ‘

Solution:

Let f( x ) = $e^{2x}$, then f ( x + h ) = e2(x + h), we know

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{[f(x + h) – f(x)]}{h}$

$\frac{d}{dx}$ f( x ) =$lim_{h\rightarrow 0}$ $\frac{[e^{2(x + h)}– e^{2x}]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{[e^{2x + 2h} – e^{2x}]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{[e^{2x} *e^{2h} – e^{2x}]}{ h}$

$\frac{d}{dx}$ f( x ) =$lim_{h\rightarrow 0}$ $\frac{e^{2x}*[e^{2h} – 1]}{h}$

$\frac{d}{dx}$ f( x ) = $e^{2x}$ * $lim_{h\rightarrow 0}$ $\frac{[e^{2h} – 1]}{h}$

Multiplying and dividing by ‘ 2 ’, we get

$\frac{d}{dx}$  f( x ) = $e^{2x} * 2 *$ $lim_{h\rightarrow 0}$ $\frac{[e^{2h} –1]}{2h}$

$\frac{d}{dx}$ f( x ) = $e^{2x}$ * 2 * ( 1 )

$\frac{d}{dx}$ f( x ) = 2$e^{2x}$

## e-5x

Find the derivative of e- 5x with respect to ‘ x ‘

Solution:

Let f( x ) = e-5x, then f(x+h) = e-5(x+h), we know

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{[f(x + h) – f(x)]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{[e^{- 5(x+h)} – e^{-5x}]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{[e^{- 5x – 5h} – e^{-5x}]}{h}$

$\frac{d}{dx}$f( x ) =$lim_{h\rightarrow 0}$ $\frac{[e^{-5x}*e^{-5h} – e^{-5x}]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$  $e^{-5x}$*$\frac{[e^{-5h} –1]}{h}$

$\frac{d}{dx}$f( x ) = $e^{-5x}$ * $lim_{h\rightarrow 0}$ $\frac{[e^{-5h}– 1]}{h}$

Multiplying and dividing by ‘ - 5 ’, we get

$\frac{d}{dx}$ f( x ) = $e^{-5x}$ *-5 * $lim_{h\rightarrow 0}$ $\frac{[e^{-5h}– 1]}{-5h}$

$\frac{d}{dx}$f( x ) = $e^{-5x}$ *- 5 * ( 1 )

$\frac{d}{dx}$f( x ) = - 5$e^{-5x}$

## e-8x

Find the derivative of e-8x with respect to ‘ x ‘

Solution:

Let f( x ) = e-8x, then f(x+h) = e-8(x+h), we know

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$  $\frac{[f(x + h) – f(x)]}{h}$

$\frac{d}{dx}$f( x ) = $lim_{h\rightarrow 0}$  $\frac{[e^{-8(x+h)} – e^{-8x}]}{ h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$  $\frac{[e^{-8x–8h} – e^{-8x}]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$  $\frac{[e^{-8x}*e^{-8h}–e^{-8x}]}{h}$

$\frac{d}{dx}$ f( x ) =  $lim_{h\rightarrow 0}$ $e^{-8x}$ * $\frac{[e^{-8h}– 1]}{ h}$

$\frac{d}{dx}$ f( x ) = $e^{-8x}$ * $lim_{h\rightarrow 0}$ $\frac{[e^{-5h} – 1]}{h}$

Multiplying and dividing by ‘ - 8 ’, we get

$\frac{d}{dx}$ f( x ) = $e^{-8x}$ *-8* $lim_{h\rightarrow 0}$ $\frac{[e^{-8h} – 1]}{-8h}$

$\frac{d}{dx}$ f( x ) = $e^{-8x}$ * - 8 * (1)

$\frac{d}{dx}$ f( x ) = - 8$e^{-8x}$

## e-sinx

Find the derivative of esin x with respect to ‘ x ‘

Solution:

Let f( x ) = esin x, then f ( x + h ) = esin ( x + h ), we know

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$  $\frac{[f(x + h) – f(x)]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$ $\frac{[e^{sin(x+h)} – e^{sin x}]}{h}$

$\frac{d}{dx}$ f( x ) = $lim_{h\rightarrow 0}$  $\frac{e^{sin x}[e^{sin(x+h)–sin x} – 1]}{h}$

Multiplying and dividing by sin (x + h) – sin x, we get

$\frac{d}{dx}$ f( x ) = $e^{sin x}$  * $lim_{h\rightarrow 0}$ $\frac{[e^{sin(x + h)– sin x} – 1]}{[sin(x + h) – sin x ]}$ * $lim_{h\rightarrow 0}$ $\frac{[sin(x + h) – sin x]}{h}$

$\frac{d}{dx}$ f( x ) = $e^{sin x}$*( 1 ) * $lim_{h\rightarrow 0}$ $\frac{[sin(x + h) – sin x]}{h}$

Using the formula of Sin C + Sin D = 2 $\sin \frac{(C –D)}{2}$ * $\cos \frac{(C + D)}{ 2}$, we get

$\frac{d}{dx}$f( x ) = $e^{sin x}$  *  $lim_{h\rightarrow 0}$ $\frac{[2 sin(\frac{h}{2}) * cos \frac{(2x + h)}{2}]}{h}$

$\frac{d}{dx}$f( x ) = $e^{sin x}$  *  $lim_{h\rightarrow 0}$  $\frac{[2 sin(h/ 2) * cos (\frac{x + h}{2})]} {2(\frac{h}{2})}$

$\frac{d}{dx}$f( x ) = $e^{sin x}$ * $lim_{h\rightarrow 0}$ $\cos$ $(\frac{x + h}{2})$ * $lim_{h\rightarrow 0}$ $[\frac{sin(\frac{h}{2})}{(\frac{h}{2})}]$
[ Since, $lim_{x\rightarrow 0}$ ($\frac{\sin x}{x}$) = 1 ]

$\frac{d}{dx}$f( x ) =  $e^{sin x}$  * cos x  * 1

$\frac{d}{dx}$f( x ) =  esin x  * cos x

## Derivative of Exponential Functions Practice Problems

Here are some practice problems for you to solve: