Is the product rule important in JEE exams?

Yes, it is frequently asked in various forms, especially when differentiating products of two or more functions.

How do you apply the chain rule in differentiation?

For a composite function y = f(g(x)), apply: (dy/dx = f'(g(x)).g'(x)

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Differentiation Formulas

Differentiation is one of the most important topics in calculus, especially for Class 12 students preparing for competitive exams like JEE. It is the process of finding the derivative of a function, which measures the rate of change of a quantity. In this blog, we will cover all differentiation formulas, including basic formulas, advanced rules, and solved examples to help you master differentiation concepts effectively.

1.0What Are Differentiation Formulas?

Differentiation formulas are predefined rules that help us compute the derivative of a function in an easy and systematic way without going back to first principles every time. These formulas are essential for solving problems efficiently in exams.

2.0Basic Differentiation Formulas (Class 12 & JEE)

Function f(x)	Differentiation Formula f'(x)
$x^{n}$	$n x^{n - 1}$
sin x	cos x
cos x	-sin x
tan x	$se c^{2} x$
cot x	$- cose c^{2} x$
sec x	sec x tan x
csc x	-csc x . cot x
$e^{x}$	$e^{x}$
$a^{x}$	$a^{x} l n a$
ln x	$\frac{1}{x}$
$l o g_{a} x$	$\frac{1}{x l n a}$

3.0Important Differentiation Formulas and Rules

1. Product Rule: $\frac{d}{d x} [u (x) \cdot v (x)] = u^{'} (x) v (x) + u (x) v^{'} (x)$

2. Quotient Rule: $\frac{d}{d x} [\frac{u ( x )}{v ( x )}] = \frac{v ( x ) u ^{'} ( x ) - u ( x ) v ^{'} ( x )}{[ v ( x ) ] ^{2}}$

3. Chain Rule: If y = f(g(x)), then: $\frac{d y}{d x} = f^{'} (g (x)) \cdot g^{'} (x)$

4.0Examples of Differentiation Formulas

Example 1: Differentiate $f (x) = x^{5} + 3 x^{3} - 2 x + 7.$

Solution: $f^{'} (x) = 5 x^{4} + 9 x^{2} - 2$

Example 2: Differentiate $f (x) = sin x \cdot cos x .$

Solution : $f^{'} (x) = (cos x) (cos x) + (sin x) (- sin x) = cos^{2} x - sin^{2} x$

Example 3: Differentiate $f (x) = sin (3 x^{2} + 2 x)$

Solution: $f^{'} (x) = cos (3 x^{2} + 2 x) \cdot (6 x + 2) = (6 x + 2) cos (3 x^{2} + 2 x)$

Example 4: Differentiate $f (x) = \frac{x ^{2} + 1}{x - 1}$

Solution:

$f^{'} (x) = \frac{( x - 1 ) ( 2 x ) - ( x ^{2} + 1 ) ( 1 )}{( x - 1 ) ^{2}} = \frac{2 x ( x - 1 ) - ( x ^{2} + 1 )}{( x - 1 ) ^{2}}$

Simplify numerator: $x^{2} - 2 x - x^{2} - 1 = x^{2} - 2 x - 1$

Thus, $f^{'} (x) = \frac{x ^{2} - 2 x - 1}{( x - 1 ) ^{2}}$

Example 5: Differentiate $f (x) = \frac{sin x}{x ^{2} + 1}$

Solution:

$f^{'} (x) = \frac{( x ^{2} + 1 ) cos x - sin x \cdot 2 x}{( x ^{2} + 1 ) ^{2}}$

So,

$f^{'} (x) = \frac{( x ^{2} + 1 ) cos x - 2 x sin x}{( x ^{2} + 1 ) ^{2}}$

Example 6: Differentiate $f (x) = cos^{3} (2 x + 1)$

Solution:

Let $u = cos (2 x + 1)$ so $f (x) = u^{3}$

Step 1 – Apply chain rule:

$f^{'} (x) = 3 u^{2} \frac{d u}{d x} = 3 cos^{2} (2 x + 1) \frac{d}{d x} [cos (2 x + 1)] .$

Step 2 – Differentiate $cos (2 x + 1)$ :

$\frac{d}{d x} [cos (2 x + 1)] = - sin (2 x + 1) \cdot 2 = - 2 sin (2 x + 1) .$

Step 3 – Combine:

$f^{'} (x) = 3 cos^{2} (2 x + 1) \cdot (- 2 sin (2 x + 1)) = - 6 cos^{2} (2 x + 1) sin (2 x + 1) .$

Example 7: Differentiate $f (x) = ln (x^{2} + s in x)$

Solution:

$f^{'} (x) = \frac{d}{d x} [ln (x^{2} + sin x)] = \frac{2 x + cos x}{x ^{2} + sin x} .$

Example 8: Differentiate $f (x) = tan^{- 1} (\frac{2 x}{1 - x ^{2}})$

Solution:

We know that: $\frac{d}{d x} [t a n^{- 1} x] = \frac{1}{1 + x ^{2}}$

Let $u = \frac{2 x}{1 - x ^{2}}$

Step 1 – Derivative of u:

$\frac{d u}{d x} = \frac{( 1 - x ^{2} ) ( 2 ) - ( 2 x ) ( - 2 x )}{( 1 - x ^{2} ) ^{2}} = \frac{2 ( 1 - x ^{2} ) + 4 x ^{2}}{( 1 - x ^{2} ) ^{2}} = \frac{2 + 2 x ^{2}}{( 1 - x ^{2} ) ^{2}}$

Step 2 – Apply chain rule:

$f^{'} (x) = \frac{1}{1 + u ^{2}} \frac{d u}{d x}$

Simplify $1 + u^{2}$ :

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

Thus, $f^{'} (x) = \frac{2 + 2 x ^{2}}{1 + 2 x ^{2} + x ^{4}}$

Example 9: Differentiate $f (x) = e^{x} s in x$

Solution :

$f^{'} (x) = \frac{d}{d x} [e^{x} sin x] = e^{x} sin x + e^{x} cos x$

So,

$f^{'} (x) = e^{x} (sin x + cos x)$

Example 10: Find $\frac{d ^{2}}{d x ^{2}} [x^{3} ln x]$

Solution:

Step 1 – First derivative:

$f^{'} (x) = \frac{d}{d x} [x^{3} ln x] = 3 x^{2} ln x + x^{3} \cdot \frac{1}{x} = 3 x^{2} ln x + x^{2}$

Step 2 – Second derivative:

$f^{''} (x) = \frac{d}{d x} [3 x^{2} ln x + x^{2}] = 3 [2 x ln x + x] + 2 x = 6 x ln x + 3 x + 2 x = 6 x ln x + 5 x$

So, $f^{''} (x) = x (6 ln x + 5)$

Example 11: Differentiate implicitly: $x^{2} + y^{2} = 25$

Solution:

Differentiate both sides w.r.t x: $2 x + 2 y \frac{d y}{d x} = 0$

Solving for $\frac{d y}{d x}$

$\frac{d y}{d x} = - \frac{x}{y}$

Example 12: Differentiate $f (x) = (x^{2} + 1)^{x}$

Solution:

Step 1 – Take logarithm on both sides: $ln f (x) = x ln (x^{2} + 1)$

Step 2 – Differentiate both sides:

$\frac{f ^{'} ( x )}{f ( x )} = ln (x^{2} + 1) + x \cdot \frac{2 x}{x ^{2} + 1} = ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1}$

Step 3 – Multiply by f(x):

$f^{'} (x) = (x^{2} + 1)^{x} [ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1}]$

5.0Practice Questions on Differentiation Formulas

Differentiate $f (x) = e^{2 x} s in x$
Find the derivative of $f (x) = l n (x^{2} + 1)$
Differentiate $f (x) = t a n^{- 1} (3 x)$
Find $\frac{d}{d x} [x^{3} l n x]$

6.0Sample Question on Differentiation Formulas

Q1. What is the basic differentiation formula for ?

Ans: $\frac{d}{d x} x^{n} = n x^{n - 1}$

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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.

Differentiation Formulas

Differentiation is one of the most important topics in calculus, especially for Class 12 students preparing for competitive exams like JEE. It is the process of finding the derivative of a function, which measures the rate of change of a quantity. In this blog, we will cover all differentiation formulas, including basic formulas, advanced rules, and solved examples to help you master differentiation concepts effectively.

1.0What Are Differentiation Formulas?

Differentiation formulas are predefined rules that help us compute the derivative of a function in an easy and systematic way without going back to first principles every time. These formulas are essential for solving problems efficiently in exams.

2.0Basic Differentiation Formulas (Class 12 & JEE)

Function f(x)	Differentiation Formula f'(x)
$x^{n}$	$n x^{n - 1}$
sin x	cos x
cos x	-sin x
tan x	$se c^{2} x$
cot x	$- cose c^{2} x$
sec x	sec x tan x
csc x	-csc x . cot x
$e^{x}$	$e^{x}$
$a^{x}$	$a^{x} l n a$
ln x	$\frac{1}{x}$
$l o g_{a} x$	$\frac{1}{x l n a}$

3.0Important Differentiation Formulas and Rules

1. Product Rule: $\frac{d}{d x} [u (x) \cdot v (x)] = u^{'} (x) v (x) + u (x) v^{'} (x)$

2. Quotient Rule: $\frac{d}{d x} [\frac{u ( x )}{v ( x )}] = \frac{v ( x ) u ^{'} ( x ) - u ( x ) v ^{'} ( x )}{[ v ( x ) ] ^{2}}$

3. Chain Rule: If y = f(g(x)), then: $\frac{d y}{d x} = f^{'} (g (x)) \cdot g^{'} (x)$

4.0Examples of Differentiation Formulas

Example 1: Differentiate $f (x) = x^{5} + 3 x^{3} - 2 x + 7.$

Solution: $f^{'} (x) = 5 x^{4} + 9 x^{2} - 2$

Example 2: Differentiate $f (x) = sin x \cdot cos x .$

Solution : $f^{'} (x) = (cos x) (cos x) + (sin x) (- sin x) = cos^{2} x - sin^{2} x$

Example 3: Differentiate $f (x) = sin (3 x^{2} + 2 x)$

Solution: $f^{'} (x) = cos (3 x^{2} + 2 x) \cdot (6 x + 2) = (6 x + 2) cos (3 x^{2} + 2 x)$

Example 4: Differentiate $f (x) = \frac{x ^{2} + 1}{x - 1}$

Solution:

$f^{'} (x) = \frac{( x - 1 ) ( 2 x ) - ( x ^{2} + 1 ) ( 1 )}{( x - 1 ) ^{2}} = \frac{2 x ( x - 1 ) - ( x ^{2} + 1 )}{( x - 1 ) ^{2}}$

Simplify numerator: $x^{2} - 2 x - x^{2} - 1 = x^{2} - 2 x - 1$

Thus, $f^{'} (x) = \frac{x ^{2} - 2 x - 1}{( x - 1 ) ^{2}}$

Example 5: Differentiate $f (x) = \frac{sin x}{x ^{2} + 1}$

Solution:

$f^{'} (x) = \frac{( x ^{2} + 1 ) cos x - sin x \cdot 2 x}{( x ^{2} + 1 ) ^{2}}$

So,

$f^{'} (x) = \frac{( x ^{2} + 1 ) cos x - 2 x sin x}{( x ^{2} + 1 ) ^{2}}$

Example 6: Differentiate $f (x) = cos^{3} (2 x + 1)$

Solution:

Let $u = cos (2 x + 1)$ so $f (x) = u^{3}$

Step 1 – Apply chain rule:

$f^{'} (x) = 3 u^{2} \frac{d u}{d x} = 3 cos^{2} (2 x + 1) \frac{d}{d x} [cos (2 x + 1)] .$

Step 2 – Differentiate $cos (2 x + 1)$ :

$\frac{d}{d x} [cos (2 x + 1)] = - sin (2 x + 1) \cdot 2 = - 2 sin (2 x + 1) .$

Step 3 – Combine:

$f^{'} (x) = 3 cos^{2} (2 x + 1) \cdot (- 2 sin (2 x + 1)) = - 6 cos^{2} (2 x + 1) sin (2 x + 1) .$

Example 7: Differentiate $f (x) = ln (x^{2} + s in x)$

Solution:

$f^{'} (x) = \frac{d}{d x} [ln (x^{2} + sin x)] = \frac{2 x + cos x}{x ^{2} + sin x} .$

Example 8: Differentiate $f (x) = tan^{- 1} (\frac{2 x}{1 - x ^{2}})$

Solution:

We know that: $\frac{d}{d x} [t a n^{- 1} x] = \frac{1}{1 + x ^{2}}$

Let $u = \frac{2 x}{1 - x ^{2}}$

Step 1 – Derivative of u:

$\frac{d u}{d x} = \frac{( 1 - x ^{2} ) ( 2 ) - ( 2 x ) ( - 2 x )}{( 1 - x ^{2} ) ^{2}} = \frac{2 ( 1 - x ^{2} ) + 4 x ^{2}}{( 1 - x ^{2} ) ^{2}} = \frac{2 + 2 x ^{2}}{( 1 - x ^{2} ) ^{2}}$

Step 2 – Apply chain rule:

$f^{'} (x) = \frac{1}{1 + u ^{2}} \frac{d u}{d x}$

Simplify $1 + u^{2}$ :

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

Thus, $f^{'} (x) = \frac{2 + 2 x ^{2}}{1 + 2 x ^{2} + x ^{4}}$

Example 9: Differentiate $f (x) = e^{x} s in x$

Solution :

$f^{'} (x) = \frac{d}{d x} [e^{x} sin x] = e^{x} sin x + e^{x} cos x$

So,

$f^{'} (x) = e^{x} (sin x + cos x)$

Example 10: Find $\frac{d ^{2}}{d x ^{2}} [x^{3} ln x]$

Solution:

Step 1 – First derivative:

$f^{'} (x) = \frac{d}{d x} [x^{3} ln x] = 3 x^{2} ln x + x^{3} \cdot \frac{1}{x} = 3 x^{2} ln x + x^{2}$

Step 2 – Second derivative:

$f^{''} (x) = \frac{d}{d x} [3 x^{2} ln x + x^{2}] = 3 [2 x ln x + x] + 2 x = 6 x ln x + 3 x + 2 x = 6 x ln x + 5 x$

So, $f^{''} (x) = x (6 ln x + 5)$

Example 11: Differentiate implicitly: $x^{2} + y^{2} = 25$

Solution:

Differentiate both sides w.r.t x: $2 x + 2 y \frac{d y}{d x} = 0$

Solving for $\frac{d y}{d x}$

$\frac{d y}{d x} = - \frac{x}{y}$

Example 12: Differentiate $f (x) = (x^{2} + 1)^{x}$

Solution:

Step 1 – Take logarithm on both sides: $ln f (x) = x ln (x^{2} + 1)$

Step 2 – Differentiate both sides:

$\frac{f ^{'} ( x )}{f ( x )} = ln (x^{2} + 1) + x \cdot \frac{2 x}{x ^{2} + 1} = ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1}$

Step 3 – Multiply by f(x):

$f^{'} (x) = (x^{2} + 1)^{x} [ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1}]$

5.0Practice Questions on Differentiation Formulas

Differentiate $f (x) = e^{2 x} s in x$
Find the derivative of $f (x) = l n (x^{2} + 1)$
Differentiate $f (x) = t a n^{- 1} (3 x)$
Find $\frac{d}{d x} [x^{3} l n x]$

6.0Sample Question on Differentiation Formulas

Q1. What is the basic differentiation formula for ?

Ans: $\frac{d}{d x} x^{n} = n x^{n - 1}$

Frequently Asked Questions

Is the product rule important in JEE exams?

How do you apply the chain rule in differentiation?

Frequently Asked Questions

Is the product rule important in JEE exams?

How do you apply the chain rule in differentiation?

Join ALLEN!

Differentiation Formulas

1.0What Are Differentiation Formulas?

2.0Basic Differentiation Formulas (Class 12 & JEE)

3.0Important Differentiation Formulas and Rules

4.0Examples of Differentiation Formulas

5.0Practice Questions on Differentiation Formulas

6.0Sample Question on Differentiation Formulas

Table of Contents

Join ALLEN!

Differentiation Formulas

1.0What Are Differentiation Formulas?

2.0Basic Differentiation Formulas (Class 12 & JEE)

3.0Important Differentiation Formulas and Rules

4.0Examples of Differentiation Formulas

5.0Practice Questions on Differentiation Formulas

6.0Sample Question on Differentiation Formulas

Table of Contents