Exponential functions can be differentiated using the chain rule. One of the most intriguing and functional characteristics of the natural exponential function is that *it is its own derivative*.

In other words, it has solution to the differential equation being the same such that,*y’* = *y*.The exponential function which has the property that the slope of the tangent line at (0,1) has the value *m*_{0} = 1 is called the **natural exponential function**. It can be written as exp(*x*) or more frequently as *e ^{x}*, where

*e*is the base of the exponential function which has a unit slope at (0,1).

Let u be a differentiable function of x, then we can accept the following properties:

- $\frac{d}{dx}\lbrack e^{x}\rbrack =e^{x}$
- $\frac{d}{dx}\lbrack e^{u}\rbrack =e^{u}\cdot \frac{du}{dx}$
- $\frac{d}{dx}\lbrack e^{f(x)}\rbrack =e^{f(x)}\cdot f'(x)$

This can be proved using the limit formula:

h | $\frac{e^{h}-1}{h}$ |

1 |
1.71828 |

0.1 |
1.05171 |

0.01 |
1.00502 |

0.001 |
1.00050 |

0.0001 |
1.00005 |

0.00001 |
1.00001 |

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

We can use the chain rule to find the derivative of the a$^{x}$.

$y=a^{x}

\ln y=\ln a^{x}=x\ln a

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

$\frac{d}{dx}a^{x}= \frac{d}{dx} (e^{x\ln a})= e^{x\ln a} \frac{d}{dx} (x\ln a)=e^{x\ln a} \cdot \ln a$

$\frac{d}{dx}a^{x}=a^{x}\cdot \ln a$

**Example: Find the derivative of $f(x)=e^{3+5x}$**

$f'(x)=e^{3+5x}\cdot \frac{d}{dx}(3+5x)$

$f'(x)=5e^{3+5x}$

Example: Differentiate the following function: y = e$^{tan x}$

Solution: Use chain rule, let u = tan x

Then, we have y = e$^{u}$.

Therefore, $\frac{dy}{dx}=\frac{dy}{dx}\cdot \frac{du}{dx} = e^{u}\frac{du}{dx}=e^{\tan x}\sec ^{2}x $

A derivative of natural logarithm function:

Let $y=\ln x$ such that x ≠ 0

Then,

$e^{y}=x$

$\frac{d}{dx}e^{y}= \frac{d}{dx} x$

$e^{y}\frac{dy}{dx}= 1$

$x\frac{dy}{dx}= 1$

$\frac{dy}{dx}=\frac{1}{x}$

$\frac{d}{dx}(\ln x)=\frac{1}{x}$

$\frac{d}{dx}(\ln f(x)=\frac{f'(x)}{f(x)};[f(x)>0]$

**Example: Differentiate y = ln(x^{3} + 1).**

Solution, Use chain rule to solve this function:

Let *u *=* x*^{3}+1, then y = ln u

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=\frac{1}{u}\cdot \frac{du}{dx}=\frac{1}{x^{3}+1}(3x^{2})=\frac{3x^{2}}{x^{3}+1}$

**Example: Find $\frac{d}{dx}(\log _{a}x)$**

Let *y* = log* _{a}x. *Then,

*a*=

^{y}*x*.

We can differentiate the equation above implicitly with respect to *x*, we obtain: $a^{y}(lna)\frac{dy}{dx}=1$

Therefore, we obtain: $\frac{dy}{dx}=\frac{1}{a^{y}\ln a}=\frac{1}{x\ln a}$

If we put *a *=* e *in the above formula, then the factor on the right side becomes ln* e *=1and we get the formula for the derivative of the natural logarithmic function log* _{e }x *=ln

*x.*

$\frac{d}{dx}(\ln x)=\frac{1}{x}$

The differentiation formula can be manipulated in the simplest form when *a *=* e *because ofln* e *=1.

**Example: Find f ’(x) if f(x) = ln |x|.**

Solution: We know:

It follows that

Thus, f’(x) = 1/x for all x not equal to 0.

Likewise, $\frac{d}{dx}\ln \vert x\vert =\frac{1}{x}$

**Example: Differentiate $f(x)= \sqrt{\ln x}$**

$f^{‘}(x)=\frac{1}{2}(\ln x)^{-\frac{1}{2}}\cdot \frac{d}{dx}(\ln x)=\frac{1}{2\sqrt{\ln x}}\cdot \frac{1}{x}=\frac{1}{2x\sqrt{\ln x}}$

**Example: Differentiate y = ln(sin x)**

$\frac{d}{dx}\ln (sinx)=\frac{1}{\sin x}\frac{d}{du}(\sin x)=\frac{1}{\sin x}\cos x=\cot x$