When should we use Logarithmic Differentiation?

Use it when dealing with complex products, quotients, or variable exponents like Xˣ, or when standard rules make differentiation tedious.

Is Logarithmic Differentiation important for JEE/Advanced?

Yes, it's frequently used in calculus problems in JEE Advanced for simplifying derivatives that are otherwise tough to handle.

Can we apply logarithmic differentiation to any function?

It’s best suited for positive-valued functions where log can be defined. Be cautious with domains.

How is this different from implicit differentiation?

Implicit differentiation is used when y is defined implicitly, while logarithmic differentiation is a technique used to simplify explicit functions before differentiating.

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate complex functions, especially those involving products, quotients, or variables in exponents. By taking the natural logarithm of both sides of an equation, it simplifies differentiation using logarithmic identities. The steps involve applying \ln, simplifying, differentiating implicitly, and solving for the derivative. It's especially useful for functions like $y = x^{x}$ or $y = \frac{( x ^{2} + 1 ) ^{3}}{x - 1}$ . This method reduces complicated expressions into simpler, more manageable forms for differentiation.

1.0What is Logarithmic Differentiation?

Logarithmic Differentiation is a technique where we take the natural logarithm (usually log base e) on both sides of a function and then differentiate implicitly. It’s particularly effective for:

Products of multiple functions
Quotients
Exponential functions like $f (x) = (x^{2} + 1)^{x}$

2.0Logarithmic Differentiation Formula

If y = f(x), then using logarithmic differentiation:

$ln y = ln f (x)$

Differentiate both sides:

$\frac{1}{y} \cdot \frac{d y}{d x} = \frac{d}{d x} [ln f (x)]$

Then,

$\frac{d y}{d x} = y \cdot \frac{d}{d x} [ln f (x)] = f (x) \cdot \frac{d}{d x} [ln f (x)]$

3.0Solved Examples of Logarithmic Differentiation

Example 1: Find the derivative of $y = x^{x}$

Solution:

$ln y = ln (x^{x}) = x ln x$

Differentiate both sides:

$\frac{1}{y} \frac{d y}{d x} = ln x + 1$

$\frac{d y}{d x} = y (ln x + 1) = x^{x} (ln x + 1)$

Example 2: Differentiate $y = \frac{( x ^{2} + 3 ) ^{4} \cdot x + 1}{( 2 x ^{2} + 1 ) ^{3}}$

Solution:

Take natural log on both sides:

$ln y = ln [\frac{( x ^{2} + 3 ) ^{4} \cdot ( x + 1 ) ^{1/2}}{( 2 x ^{2} + 1 ) ^{3}}]$

Apply log laws:

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

Differentiate:

$\frac{1}{y} \frac{d y}{d x} = \frac{4 \cdot 2 x}{x ^{2} + 3} + \frac{1}{2 ( x + 1 )} - \frac{3 \cdot 4 x}{2 x ^{2} + 1} \frac{d y}{d x} = y [\frac{8 x}{x ^{2} + 3} + \frac{1}{2 ( x + 1 )} - \frac{12 x}{2 x ^{2} + 1}]$

Substitute original y back:

$\frac{d y}{d x} = \frac{( x ^{2} + 3 ) ^{4} \cdot x + 1}{( 2 x ^{2} + 1 ) ^{3}} [\frac{8 x}{x ^{2} + 3} + \frac{1}{2 ( x + 1 )} - \frac{12 x}{2 x ^{2} + 1}]$

Example 3: Differentiate $y = \frac{( x ^{2} + 1 ) ^{3}}{x - 1}$

Solution:

Take natural log:

$ln y = ln (\frac{( x ^{2} + 1 ) ^{3}}{( x - 1 ) ^{1/2}}) = 3 ln (x^{2} + 1) - \frac{1}{2} ln (x - 1)$

Differentiate both sides:

$\frac{1}{y} \frac{d y}{d x} = \frac{3 \cdot 2 x}{x ^{2} + 1} - \frac{1}{2 ( x - 1 )}$

Multiply by yy:

$\frac{d y}{d x} = \frac{( x ^{2} + 1 ) ^{3}}{x - 1} (\frac{6 x}{x ^{2} + 1} - \frac{1}{2 ( x - 1 )})$

Example 4: Find the derivative of y wrt x. $y = (sin x)^{t a n x}$

Solution:

Take logarithm:

$ln y = tan x \cdot ln (sin x)$

Differentiate:

$\frac{1}{y} \frac{d y}{d x} = sec^{2} x \cdot ln (sin x) + tan x \cdot \frac{c o s x}{s i n x} = sec^{2} x \cdot ln (sin x) + tan x \cdot cot x = sec^{2} x \cdot ln (sin x) + 1$

So,

$\frac{d y}{d x} = (sin x)^{t a n x} (sec^{2} x \cdot ln (sin x) + 1)$

Example 5: Differentiate $y = (x^{2} + 3 x + 2)^{x^{2} + 1}$

Solution:

Step 1: Take natural log:

$ln y = x^{2} + 1 \cdot ln (x^{2} + 3 x + 2)$

Step 2: Differentiate:

$\frac{1}{y} \frac{d y}{d x} = \frac{x}{x ^{2} + 1} ln (x^{2} + 3 x + 2) + x^{2} + 1 \cdot \frac{2 x + 3}{x ^{2} + 3 x + 2}$

Final Answer:

$\frac{d y}{d x} = (x^{2} + 3 x + 2)^{x^{2} + 1} [\frac{x l n ( x ^{2} + 3 x + 2 )}{x ^{2} + 1} + x^{2} + 1 \cdot \frac{2 x + 3}{x ^{2} + 3 x + 2}]$

Example 6: Differentiate $y = \frac{[ l n x ] ^{x}}{x ^{l n x}}$

Solution:

Take natural log:

$ln y = x ln (ln x) - ln x \cdot ln x = x ln (ln x) - (ln x)^{2}$

Differentiate:

$\frac{1}{y} \frac{d y}{d x} = ln (ln x) + \frac{x}{l n x} \cdot \frac{1}{x} - 2 ln x \cdot \frac{1}{x}$

Simplify:

$\frac{d y}{d x} = \frac{[ l n x ] ^{x}}{x ^{l n x}} [ln (ln x) + \frac{1}{l n x} - \frac{2 l n x}{x}]$

Example 7: Differentiate $y = (\frac{( x ^{2} + 1 ) ^{5} \cdot t a n x}{e ^{x} \cdot l n x})^{x}$

Solution:

Take logarithm:

$ln y = x \cdot ln (\frac{( x ^{2} + 1 ) ^{5} \cdot t a n x}{e ^{x} \cdot l n x})$

Break it down using log rules:

$ln y = x [5 ln (x^{2} + 1) + \frac{1}{2} ln (tan x) - x - ln (ln x)]$

Differentiate using product rule:

$\frac{1}{y} \frac{d y}{d x} = [5 ln (x^{2} + 1) + \frac{1}{2} ln (tan x) - x - ln (ln x)] + x [\frac{10 x}{x ^{2} + 1} + \frac{1}{2} \frac{1}{t a n x} sec^{2} x - 1 - \frac{1}{l n x} \cdot \frac{1}{x}]$

You can now write:

$\frac{d y}{d x} = y \cdot (expression above)$

4.0Logarithmic Differentiation Practice Questions

Differentiate: $y = \frac{( 3 x ^{2} + 5 ) ^{2}}{x ^{2} + 1}$
Differentiate: $y = x^{x} \cdot x + 2$
Differentiate: $y = \frac{( x ^{2} + 1 ) ^{3} \cdot ( x + 1 ) ^{2}}{2 x ^{3} + 4}$
Find if $\frac{d y}{d x} if y = (\frac{x ^{2} + 3}{x ^{2} - 1})^{x}$
Let $y = \frac{( x ^{3} + 1 ) ^{5} \cdot ( x ^{2} - 4 ) ^{3}}{( x ^{2} + 2 x + 2 ) ^{4} \cdot x - 2}$ _.Find using logarithmic differentiation.
Differentiate: $\frac{d y}{d x}$
Differentiate: $y = \frac{x ^{3} x ^{2} + 1}{( x - 1 ) ^{2}}$
Differentiate: $y = (x^{2} + 4)^{x} \cdot e^{x l n x}$

Also Read:

Differential Equations	Derivative Function Calculus	Differentiability
First Derivative Test	Continuity and Differentiability	Integrals of Particular Functions
Differentiation and Integration	First order Differential Equation	Logarithmic Functions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate complex functions, especially those involving products, quotients, or variables in exponents. By taking the natural logarithm of both sides of an equation, it simplifies differentiation using logarithmic identities. The steps involve applying \ln, simplifying, differentiating implicitly, and solving for the derivative. It's especially useful for functions like $y = x^{x}$ or $y = \frac{( x ^{2} + 1 ) ^{3}}{x - 1}$ . This method reduces complicated expressions into simpler, more manageable forms for differentiation.

1.0What is Logarithmic Differentiation?

Logarithmic Differentiation is a technique where we take the natural logarithm (usually log base e) on both sides of a function and then differentiate implicitly. It’s particularly effective for:

Products of multiple functions
Quotients
Exponential functions like $f (x) = (x^{2} + 1)^{x}$

2.0Logarithmic Differentiation Formula

If y = f(x), then using logarithmic differentiation:

$ln y = ln f (x)$

Differentiate both sides:

$\frac{1}{y} \cdot \frac{d y}{d x} = \frac{d}{d x} [ln f (x)]$

Then,

$\frac{d y}{d x} = y \cdot \frac{d}{d x} [ln f (x)] = f (x) \cdot \frac{d}{d x} [ln f (x)]$

3.0Solved Examples of Logarithmic Differentiation

Example 1: Find the derivative of $y = x^{x}$

Solution:

$ln y = ln (x^{x}) = x ln x$

Differentiate both sides:

$\frac{1}{y} \frac{d y}{d x} = ln x + 1$

$\frac{d y}{d x} = y (ln x + 1) = x^{x} (ln x + 1)$

Example 2: Differentiate $y = \frac{( x ^{2} + 3 ) ^{4} \cdot x + 1}{( 2 x ^{2} + 1 ) ^{3}}$

Solution:

Take natural log on both sides:

$ln y = ln [\frac{( x ^{2} + 3 ) ^{4} \cdot ( x + 1 ) ^{1/2}}{( 2 x ^{2} + 1 ) ^{3}}]$

Apply log laws:

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

Differentiate:

$\frac{1}{y} \frac{d y}{d x} = \frac{4 \cdot 2 x}{x ^{2} + 3} + \frac{1}{2 ( x + 1 )} - \frac{3 \cdot 4 x}{2 x ^{2} + 1} \frac{d y}{d x} = y [\frac{8 x}{x ^{2} + 3} + \frac{1}{2 ( x + 1 )} - \frac{12 x}{2 x ^{2} + 1}]$

Substitute original y back:

$\frac{d y}{d x} = \frac{( x ^{2} + 3 ) ^{4} \cdot x + 1}{( 2 x ^{2} + 1 ) ^{3}} [\frac{8 x}{x ^{2} + 3} + \frac{1}{2 ( x + 1 )} - \frac{12 x}{2 x ^{2} + 1}]$

Example 3: Differentiate $y = \frac{( x ^{2} + 1 ) ^{3}}{x - 1}$

Solution:

Take natural log:

$ln y = ln (\frac{( x ^{2} + 1 ) ^{3}}{( x - 1 ) ^{1/2}}) = 3 ln (x^{2} + 1) - \frac{1}{2} ln (x - 1)$

Differentiate both sides:

$\frac{1}{y} \frac{d y}{d x} = \frac{3 \cdot 2 x}{x ^{2} + 1} - \frac{1}{2 ( x - 1 )}$

Multiply by yy:

$\frac{d y}{d x} = \frac{( x ^{2} + 1 ) ^{3}}{x - 1} (\frac{6 x}{x ^{2} + 1} - \frac{1}{2 ( x - 1 )})$

Example 4: Find the derivative of y wrt x. $y = (sin x)^{t a n x}$

Solution:

Take logarithm:

$ln y = tan x \cdot ln (sin x)$

Differentiate:

$\frac{1}{y} \frac{d y}{d x} = sec^{2} x \cdot ln (sin x) + tan x \cdot \frac{c o s x}{s i n x} = sec^{2} x \cdot ln (sin x) + tan x \cdot cot x = sec^{2} x \cdot ln (sin x) + 1$

So,

$\frac{d y}{d x} = (sin x)^{t a n x} (sec^{2} x \cdot ln (sin x) + 1)$

Example 5: Differentiate $y = (x^{2} + 3 x + 2)^{x^{2} + 1}$

Solution:

Step 1: Take natural log:

$ln y = x^{2} + 1 \cdot ln (x^{2} + 3 x + 2)$

Step 2: Differentiate:

$\frac{1}{y} \frac{d y}{d x} = \frac{x}{x ^{2} + 1} ln (x^{2} + 3 x + 2) + x^{2} + 1 \cdot \frac{2 x + 3}{x ^{2} + 3 x + 2}$

Final Answer:

$\frac{d y}{d x} = (x^{2} + 3 x + 2)^{x^{2} + 1} [\frac{x l n ( x ^{2} + 3 x + 2 )}{x ^{2} + 1} + x^{2} + 1 \cdot \frac{2 x + 3}{x ^{2} + 3 x + 2}]$

Example 6: Differentiate $y = \frac{[ l n x ] ^{x}}{x ^{l n x}}$

Solution:

Take natural log:

$ln y = x ln (ln x) - ln x \cdot ln x = x ln (ln x) - (ln x)^{2}$

Differentiate:

$\frac{1}{y} \frac{d y}{d x} = ln (ln x) + \frac{x}{l n x} \cdot \frac{1}{x} - 2 ln x \cdot \frac{1}{x}$

Simplify:

$\frac{d y}{d x} = \frac{[ l n x ] ^{x}}{x ^{l n x}} [ln (ln x) + \frac{1}{l n x} - \frac{2 l n x}{x}]$

Example 7: Differentiate $y = (\frac{( x ^{2} + 1 ) ^{5} \cdot t a n x}{e ^{x} \cdot l n x})^{x}$

Solution:

Take logarithm:

$ln y = x \cdot ln (\frac{( x ^{2} + 1 ) ^{5} \cdot t a n x}{e ^{x} \cdot l n x})$

Break it down using log rules:

$ln y = x [5 ln (x^{2} + 1) + \frac{1}{2} ln (tan x) - x - ln (ln x)]$

Differentiate using product rule:

$\frac{1}{y} \frac{d y}{d x} = [5 ln (x^{2} + 1) + \frac{1}{2} ln (tan x) - x - ln (ln x)] + x [\frac{10 x}{x ^{2} + 1} + \frac{1}{2} \frac{1}{t a n x} sec^{2} x - 1 - \frac{1}{l n x} \cdot \frac{1}{x}]$

You can now write:

$\frac{d y}{d x} = y \cdot (expression above)$

4.0Logarithmic Differentiation Practice Questions

Differentiate: $y = \frac{( 3 x ^{2} + 5 ) ^{2}}{x ^{2} + 1}$
Differentiate: $y = x^{x} \cdot x + 2$
Differentiate: $y = \frac{( x ^{2} + 1 ) ^{3} \cdot ( x + 1 ) ^{2}}{2 x ^{3} + 4}$
Find if $\frac{d y}{d x} if y = (\frac{x ^{2} + 3}{x ^{2} - 1})^{x}$
Let $y = \frac{( x ^{3} + 1 ) ^{5} \cdot ( x ^{2} - 4 ) ^{3}}{( x ^{2} + 2 x + 2 ) ^{4} \cdot x - 2}$ _.Find using logarithmic differentiation.
Differentiate: $\frac{d y}{d x}$
Differentiate: $y = \frac{x ^{3} x ^{2} + 1}{( x - 1 ) ^{2}}$
Differentiate: $y = (x^{2} + 4)^{x} \cdot e^{x l n x}$

Also Read:

Differential Equations	Derivative Function Calculus	Differentiability
First Derivative Test	Continuity and Differentiability	Integrals of Particular Functions
Differentiation and Integration	First order Differential Equation	Logarithmic Functions

Frequently Asked Questions

When should we use Logarithmic Differentiation?

Is Logarithmic Differentiation important for JEE/Advanced?

Can we apply logarithmic differentiation to any function?

How is this different from implicit differentiation?

Frequently Asked Questions

When should we use Logarithmic Differentiation?

Is Logarithmic Differentiation important for JEE/Advanced?

Can we apply logarithmic differentiation to any function?

How is this different from implicit differentiation?

Join ALLEN!

Logarithmic Differentiation

1.0What is Logarithmic Differentiation?

2.0Logarithmic Differentiation Formula

3.0Solved Examples of Logarithmic Differentiation

4.0Logarithmic Differentiation Practice Questions

Table of Contents

Join ALLEN!

Logarithmic Differentiation

1.0What is Logarithmic Differentiation?

2.0Logarithmic Differentiation Formula

3.0Solved Examples of Logarithmic Differentiation

4.0Logarithmic Differentiation Practice Questions

Table of Contents