We will now turn our attention to simple exponential function. In general, the derivative of the exponential function $f(x)$ = $a^{x}$ is given by $f^{'}(x)$ = $a^{x}\ln a$. A special case of the rule for differentiating exponential function occurs when the base is $e$. The derivative of the function $f(x)$ = $e^{x}$ is $f^{'}(x)$ = $e^{x}\ln e$ = $e^{x}$. The derivative of the function $f(x)$ = $e^{x}$ is just itself.

One of the most important characteristics of the the natural exponential functions is that 'it is its own derivative.'
Let $u$ be a differential function of $x$, then Derivative of the natural exponential function can be written as

$\frac{\mathrm{d} e^{x}}{\mathrm{d} x}$ = $e^{x}$

$\frac{\mathrm{d} e^{ax}}{\mathrm{d} x}$ = $ae^{ax}$

$\frac{\mathrm{d} e^{u}}{\mathrm{d} x}$ = $e^{u}$ $\frac{\mathrm{d} u}{\mathrm{d} x}$

Proof:
lets consider $\ln e^{x}$ = $x$,

Differentiate each side of the equation with respect to $x$,

$\ln e^{x}$ = $x$

$\frac{\mathrm{d} (\ln e^{x})}{\mathrm{d} x}$ = $\frac{\mathrm{d} (x)}{\mathrm{d} x}$

$\frac{1}{e^{x}}\frac{\mathrm{d} (e^{x})}{\mathrm{d} x}$ = $1$

$\frac{\mathrm{d} (e^{x})}{\mathrm{d} x}$ = $e^{x}$
Example: The following are the example for derivative of exponential function,

1. $\frac{\mathrm{d} (e^{-x})}{\mathrm{d} x}$ = $-e^{-x}$

2. $\frac{\mathrm{d} (e^{2x})}{\mathrm{d} x}$ = $2e^{2x}$

3. $\frac{\mathrm{d} (e^{x^{2}})}{\mathrm{d} x}$ = $2x e^{x^{2}}$

We can interpret this theorem geometrically by saying that the slope of the graph of $f(x)$ = $e^{x}$ at any point $(x,e^{x})$ is equal to the y- coordinates of the point.
The following are the problems of derivative of exponential function

Solved Examples

Question 1: Find the derivative of the function $f(x)$ = $3e^{x}$.
Solution:
$f(x)$ = $3e^{x}$

$f^{'}(x)$ = $3$$\frac{\mathrm{d} (e^{x})}{\mathrm{d} x}$ 

$f^{'}(x)$ = $3e^{x}$


Question 2: Find the derivative of the function $f(x)$ = $\sin x – e^{x}$.
Solution:
$f(x)$ = $\sin x – e^{x}$

$f^{'}(x)$ = $\frac{\mathrm{d} (\sin x – e^{x})}{\mathrm{d} x}$

= $\frac{\mathrm{d} (\sin x)}{\mathrm{d} x}$ - $\frac{\mathrm{d} ( e^{x})}{\mathrm{d} x}$

=$\cos x – e^{x}$


Question 3: What is the derivative of $y$ = $e^{\frac{2x^{2}}{5}}$ ?
Solution:
Let $\frac{2x^{2}}{5}$ = $u$

then $\frac{\mathrm{d} u}{\mathrm{d} x}$ = $\frac{2}{5}$$\frac{\mathrm{d} x^{2}}{\mathrm{d} x}$

= $\frac{2}{5}$ $. 2x$ = $\frac{4}{5}$$x$  

Now $y$ = $e^{u}$ and $\frac{\mathrm{d} y}{\mathrm{d} u}$ = $e^{u}$

using the chain rule 

$\therefore$ $\frac{\mathrm{d} y}{\mathrm{d} x}$ = $\frac{\mathrm{d} y}{\mathrm{d} u}. \frac{\mathrm{d} u}{\mathrm{d} x}$

= $e^{u}\frac{4}{5}x$ 

= $\frac{4x}{5}$$ e^{\frac{2x^{2}}{5}}$


Question 4: If $x^{y}$ = $e^{x-y}$, show that
 $\frac{\mathrm{d} y}{\mathrm{d} x}$
= $\frac{\log x}{(1+ \log x)^{2}}$
Solution:
$x^{y}$ = $e^{x-y}$ 

Take log on both sides, 

$y \log x$ = $(x-y)\log e$ = $(x-y)$

$y(1+\log x)$ = $x$

$y$ = $\frac{x}{1+ \log x}$

$\frac{\mathrm{d} y}{\mathrm{d} x}$ = $\frac{(1+\log x).1-x.\frac{1}{x}}{(1+\log x)^{2}}$ = $\frac{\log x}{(1+ \log x)^{2}}$


Question 5: Solve the derivative of given exponential function $\frac{2e^{z}}{1-e^{z}}$
Solution:
Let $f(z)$ = $\frac{2e^{z}}{1-e^{z}}$ 

$f^{'}(z)$ = $\frac{\mathrm{d} \frac{2e^{z}}{1-e^{z}}}{\mathrm{d} z}$

let $u$ = $2e^{z}$ and $v$ = $1- e^{z}$

$\frac{\mathrm{d} u}{\mathrm{d} z}$ = $2e^{z}$

$\frac{\mathrm{d} v}{\mathrm{d} z}$ = $0-e^{z}$

by using quotient rule we have 

$\frac{\mathrm{d} \frac{u}{v}}{\mathrm{d} z}$ = $\frac{v \frac{\mathrm{d} u}{\mathrm{d} z}-u\frac{\mathrm{d} v}{\mathrm{d} z}}{v^{2}}$
 
$2$$\frac{\mathrm{d}(\frac{e^{z}}{1-e^{z}})}{\mathrm{d} z}$ = $\frac{1-e^{z}2e^{z}-2e^{z}(0-e^{z})}{(1-e^{z})^{2}}$

= $\frac{2e^{z}- 2(e^{z})^{2}+2(e^{z})^{2}}{(1-e^{z})^{2}}$

= $\frac{2e^{z}}{(1-e^{z})^{2}}$