If the given function if a natural logarithmic function say,
f(x) = ln (x)
then ƒ’(x) = $\frac{1}{x}$

Proof:

Let y = ln (x)
then by applying definition of logarithm,
ey = x

Take derivative of both sides with respect to x

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

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

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

If the function is a composite function such that ƒ(u) = ln u

and u(x) is some function in x then its derivative can be computed by applying chain rule.

ƒ (u) = ln (u)

ƒ‘ (u) = $\frac{1}{u}$

Since u is a function of x we can write

ƒ’(u(x)) = $\frac{1}{u}$ u’

Solved Examples

Question 1: f(x) = ln (3x - 5)
Solution:
 
f(x) = ln (3x - 5)

ƒ‘(x) = $\frac{1}{(3x-5)}$ $\frac{d}{dx}$ (3x - 5)

ƒ‘(x) = $\frac{1}{(3x-5)}$ (3-0)

ƒ‘(x) = $\frac{3}{(3x-5)}$
 

Question 2: ƒ(x) = ln $[(2+x) (1+x^2 )^2 (3 + x^3)^3]$
Solution:
 
ƒ(x) = ln $[(2+x) (1+x^2 )^2 (3 + x^3)^3]$

First simplify the log as sum of terms.

ƒ(x) = ln(2+x) + ln $(1+x^2)^2$ + ln $(3+x^3)^3$

ƒ(x) = ln(2+x) + 2 ln $(1+x^2)$ + 3ln$(3 + x^3)$

Take the derivative of the function
ƒ’(x) = $\frac{1}{(2+x)}$ $\frac{d}{dx}$ (2+x)+2 $\times$ $\frac{1}{(1+x^2)}$ $\frac{d (1+x^2)}{dx}$ + 3 $\times$ $\frac{1}{(3+x^3)}$ $\frac{d}{dx}$ $(3+x^3)$

= $\frac{1}{(2+x)}$ (0+1) + 2 $\times$ $\frac{1}{(1+x^2)}$ (0+2x) + 3 $\times$ $\frac{1}{(3+x^3)}$ $(0+3x^2)$

ƒ‘( x) = $\frac{1}{(2+x)}$ + $\frac{4x}{(1+x^2)}$ + $\frac{9x^2}{(3+x^3)}$
 

Derivative of log with base other than e:

ƒ(x) = $log_bx$

then ƒ‘(x) = $\frac{1}{x ln(b)}$

Proof:

ƒ(x) = $log_bx$

Apply the rule for change of base

f(x) = $\frac{log_ex}{log_eb}$

since $log_ex$ is commonly written as ln x we have

F(x) = $\frac{ln\ x}{ln\ b}$

where ln b is a constant term

so f‘(x) = $\frac{1}{ln\ b}$ $\times$ $\frac{d}{dx}$ ln x

ƒ‘(x) = $\frac{1}{ln\ b}$ $\times$ $\frac{1}{x}$

For derivative of a logarithm with base 2 we have

ƒ(x) = $log_2x$

ƒ‘(x) = $\frac{1}{ln\ 2}$ $\times$ $\frac{1}{x}$

= $\frac{1}{0.6931 x}$

Solved Example

Question: Find the derivative of the function
ƒ(x) = $log_2$ $x(3x^2 - 4x)$
Solution:
 
Here base of the log is 2.

ƒ(x) = $log_2x$ + $log_2$ $(3x^2 - 4x)$

ƒ’(x) = $\frac{d}{dx}$ $log_2x$ + $\frac{d}{dx}$ $log_2$ $(3x^2-4x)$

ƒ‘(x) = $\frac{1}{ln\ 2}$ $\times$ $\frac{1}{x}$ + $\frac{1}{ln\ 2}$ $\times$ $\frac{1}{3x^2 - 4x}$ $\times$ (3 $\times$ 2x - 4)

ƒ‘(x) = $\frac{1}{ln\ 2}$ $\times$ $\frac{1}{x}$ + $\frac{1}{ln\ 2}$ $\times$ $\frac{(6x - 4)}{3x^2 - 4x}$

ƒ’(x) = $\frac{1}{ln\ 2}$ ($\frac{1}{x}$ + $\frac{6x - 4}{3x^2 - 4x}$)

ƒ‘(x) = $\frac{1}{0.6931}$ ($\frac{1}{x}$ + $\frac{6x - 4}{3x^2 - 4x}$)
 

Let the base of the log be 5 then the function will be like

ƒ(x) = $log_5\ x$

Derivative for this function will be

ƒ‘(x) = $\frac{1}{x\ ln\ 5}$

ƒ‘(x) = $\frac{1}{x \times 1.6094}$

Solved Example

Question: ƒ(t) = $log_5$$(6t^4 + 3t^2)$
Solution:
 
ƒ(t) = $log_5$$(6t^4 + 3t^2)$

ƒ’(t) = $\frac{d}{dt}$ $log_5$$(6t^4 + 3t^2)$

ƒ‘(t) = $\frac{1}{(ln\ 5(6t^4 + 3t^2)}$ $\times$ $(6 \times 4 t^3 + 3 \times 2t)$

ƒ‘(t) = $\frac{(24t^3 + 6t)}{ln\ 5(6t^4 + 3t^2)}$
 

A logarithmic function with base 10 is known as a common logarithm.
the function is written as

ƒ(x) = log x

When the base is 10 ,base is not shown and logarithm is written with the word “log “.

Derivative of the function will be

ƒ’(x) = $\frac{1}{xln\ 10}$

Solved Example

Question: Find the derivative of the function

ƒ(x) = $log_{10} \sqrt{x+2}$ (sin$x$)
Solution:
 
Since the function is written with log without the base shown ,it is assumed that the base is 10

ƒ(x) = $log_{10} \sqrt{x+2}$ (sin$x$)

ƒ(x) = $log_{10} \sqrt{x+2}$ + $log_{10}$ sin$x$

ƒ(x) = $\frac{1}{2}$ log(x + 2) + log$sinx$

ƒ‘(x) = $\frac{1}{2ln10}$ $\times$ $\frac{1}{x + 2}$ + $\frac{1}{ln10}$ $\times$ $\frac{1}{sinx}$ $\times$ cos$x$

ƒ‘(x) = $\frac{1}{2(x + 2)ln10}$ + $\frac{cosx}{sin\ x\ ln10}$
 


Solved Examples

Question 1: f(x) = ln($x$^2$ $sin$^2$ x)
Solution:
 
First simplify the logarithm.

ƒ(x) = ln $x$^2$ + ln $sin$^2$ x

ƒ(x) = 2 ln x + 2 ln sin x

Differentiate the function with respect to x

ƒ(x) = 2 $\times$ $\frac{1}{x}$ + 2 $\times$ $\frac{1}{sinx}$ $\times$ cos$x$

ƒ‘(x) = $\frac{2}{x}$ + 2 cotx
 

Question 2: Find derivative of
y = $xln^3$ x
Solution:
 
y = $xln^3$ x
means

y = $x(lnx)^3$

Please note that we can not apply the rule log an = n log a since $(ln\ x)^3$ is not same as ln ($x^3$) .

Let us find the derivative for this function. Since this is a composite function we will use chain rule.

$\frac{dy}{dx}$ = x $\times$ [3(lnx)$^2$] $\times$ $\frac{d}{dx}$ (ln x) + (ln x)$^3$ $\frac{d}{dx}$ x

$\frac{dy}{dx}$ = x $\times$[3(lnx)$^2$] $\times$ $\frac{1}{x}$ +(lnx)$^3$ $\times$ 1

= [3(lnx)$^2$] + (lnx)$^3$

y’ = (ln x)$^2$ (3 + ln x)
 

Question 3: Find the derivative of
4 ln xy + cos y = x$^3$
Solution:
 
First simplify the function using rule of sum of logs

4(ln x + ln y) + cos y = x$^3$

Take derivative of both the sides. This is an implicit function:

4 ($\frac{1}{x}$ + $\frac{1}{y}$ $\frac{dy}{dx}$) + (-siny)$\frac{dy}{dx}$ = 3x$^2$

Collect the terms with $\frac{dy}{dx}$ on one side we get

$\frac{4}{y}$$\frac{dy}{dx}$ - siny $\frac{dy}{dx}$ = 3x$^2$ -$\frac{4}{x}$

$\frac{dy}{dx}$($\frac{4}{y}$ - siny) = 3x$^2$ -$\frac{4}{x}$

$\frac{dy}{dx}$ = $\frac{(3x^2 - \frac{4}{x})}{(\frac{4}{y} - siny)}$

$\frac{dy}{dx}$ = $\frac{(3x^3 y - 4y)}{4x - xy\ sin\ y}$

y’ = $\frac{(3x^3 y - 4y)}{4x - xy\ sin\ y}$
 

Question 4: Find derivative of y = √(x) ex2 (x$^2$ + 6)$^8$
Solution:
 
This function is a product of three functions √(x), ex2 and (x$^2$ + 6)$^8$
so if the derivative is carried out directly then we will need to apply product rule twice. Moreover the functions, ex2 and (x$^2$ + 6)$^8$ are composite functions with functions inside another function ,which requires chain rule.

To make the problem simple and easier, logarithm can be applied. Take natural log of both sides, which helps convert the products into sum functions and eliminate the powers as well.

y = √x ex2 (x$^2$ + 6)$^8$

Take natural log of both sides

ln y = ln (√x ex2 (x$^2$ + 6)$^8$)

Simplify the product using laws of sum of logs.

ln y = ln(√x) + ln ex2 + ln(x$^2$ + 6)$^8$

Apply the law of power of logs(ln a$^n$ = n ln a)

= $\frac{1}{2}$ lnx + x$^2$ ln$e$ + 8 ln(x$^2$ + 6)

Plug in ln e =1.

ln y = $\frac{1}{2}$lnx + x$^2$ + 8 ln(x$^2$ + 6)

We will apply implicit differentiation here. Take derivative of both sides:

$\frac{1}{y}$y' = $\frac{1}{2}$ $\times$ $\frac{1}{x}$ + 2x + $\frac{8}{x^2 + 6}$ $\times$ $\frac{d}{dx}$ (x$^2$ + 6)

$\frac{y^'}{y}$ = $\frac{1}{2x}$ + 2x + $\frac{8(2x)}{x^2 + 6}$

$\frac{y^'}{y}$ = $\frac{1}{2x}$ + 2x + $\frac{16x}{x^2 + 6}$

Since we need y’, let us multiply both sides by y

y^'= y $\times$ ($\frac{1}{2x}$ + 2x + $\frac{16x}{x^2 + 6}$)

y’ =√x ex2 (x$^2$ + 6)$^8$ ($\frac{1}{2x}$ + 2x + $\frac{6x}{x^2 + 6}$)
 

Question 5: Find derivative of the function:

y = 2sin2 x
Solution:
 
Take log of both sides with base 2 since base of the exponent on right side is 2.

$log_2y$ = $log_2$2sin2 x

$log_2y$ = $sin^2$x $log_2$2

Since by the rule $log_a$a = 1 we will have $log_2$2 = 1

$log_2$y = $sin^2$x

Take derivative of both sides, apply the implicit differentiation

$\frac{1}{yln\ 2}$ $y^'$ = 2 sinx (cosx)

Multiply both sides by yln 2 to get y’

$y^'$ = y ln2 (2 sinx cosx)
= (ln 2) 2sin2 x (sin2x)

[Note that 2sinx cos x = sin(2x)]

So

y’= 0.6931 2sin2 x x (sin2x)