Derivatives of Exponential Functions

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

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