Differentiation Questions: Class 12 Maths Notes, Definition, Formula, Properties & Solved Examples

Updated on Jun 17, 2025 15:15 IST

Differentiation is a fundamental concept in calculus that is used to find the derivatives of functions. This concept is used in mathematics, economics, Physics, engineering, and many other fields. The differentiation of a capacity f(x) is spoken to as f'(x). If f(x) = y, at that point f'(x) = dy/dx, which implies y is separated as for x. Before we begin tackling a few inquiries dependent on differentiation, let us see the overall differentiation equations utilized here.

Function f(x) = y	Differentiation of function f’(x) = dy/dx
xⁿ	nx^n-1
e^x	e^x
ln(x)	1/x
sin x	cos x
cos x	-sin x
tan x	sec²x
K (constant)	0

Students can practice the differentiation questions to prepare for the exam to obtain good marks. Also, students will get familiar with the exam difficulty level. The experts at Siksha have updated the NCERT Solutions for exam preparation.

Table of content

What are Differential Questions?
Weightage of Differential Questions
Illustrated Examples on Differential Questions
FAQs on Differential Questions

What are Differential Questions?

1. Differentiate y = (tan x - 1)/ sec x
y = (tan x - 1)/ sec x
u = tan x - 1 ===> u' = sec² x - 0
v = sec x ===> v' = sec x tan x

2.Differentiate y = sin x / x²

y = sin x / x²

u = sin x ==> u' = cos x

u = x2 ==> v' = 2 x

dy/dx = [x² (cos x) - sin x(2x)]/(x²)²

dy/dx = [x² (cos x) - 2x sin x]/x⁴

dy/dx = x [x cos x - 2 sin x]/x⁴

dy/dx = [x cos x - 2 sin x]/x³

3. Differentiate y = tan θ (sin θ + cos θ)
y = tan θ (sin θ + cos θ)
u = tan θ ==> u' = sec² θ
v = sin θ + cos θ ==> v' = cos θ - sin θ
dy/dx = tan θ (cos θ - sin θ) + (sin θ + cos θ) (sec²θ)
= tan θ cos θ - tan θ sin θ + sin θ sec²θ + cos θ sec²θ

4. Differentiate y = cosec x ⋅ cot x
y = cosec x ⋅ cot x
u = cosec x ==> u' = -cosec x cot x
v = cot x ==> v' = -cosec² x
dy/dx = cosec x (-cosec² x) + cot x(-cosec x cot x)
= - cosec³ x - cosec x cot² x
= - (1/sin³ x) - (1/sin x) (cos² x/sin² x)
= - (1/sin³ x) - (cos² x/sin³ x)
dy/dx = - (1 + cos²x) /sin³ x

5. Differentiate 10x² with respect to x.
y = 10x²
y’ = d(10x²)/dx
y’ = 2.10.x = 20x
Therefore, d(10x2)/dx = 20 x

6. Compute the derivative of f(x) = sin²x.
f(x) = sin²x = sin x sin x
= d(sin x)/dx. sin x + sin x.d(sin x)/dx
= cos x. sin x + sin x cos x
= 2 sin x cos x

7. Differentiate tan²x.
Say, y = tan²x
dy/dx = d(tan²x)/dx
= 2tan^2-1x. d(tan x)/dx
= 2tan x sec²x
d[tan²x]/dx = 2tan x sec²x

Weightage of Differential Questions

In Class 11: The topic has been discussed with basic explanations and the ways to solve the particular question. It has a weightage of 30 Marks. Basic concepts of differentiation are covered in the Continuity and Differentiability Class 12 Maths Chapter, including the concept of differentiability and its use to define whether a function is differentiable or not.

Also Read: NCERT Solution Class 12 Maths | Class 11 Maths NCERT Solutions

Illustrated Examples on Differential Questions

1. Differentiate sin(3x+5)

Solution.
Say, y = sin (3x+5)
dy/dx = d[sin(3x+5)]/dx
= cos (3x+5) d(3x+5)/dx [ By chain rule]
= cos (3x+5) [3]
y’ = 3 cos (3x+5)
d[sin(3x+5)]/dx = 3 cos (3x+5)

2.Differentiate 20x-4 + 9.

Solution.
y = 20x-4 + 9
y’ = d(20x-4 + 9)/dx
y’ = d(20x-4)/dx + d(9)/dx
y’ = -4.20.x-4-1+0
y’ = -80x-5
Therefore, d(20x-4 + 9)/dx = -80x-5
3. Differentiate x⁵ with respect to x.

Solution.

Given, y = x⁵
On differentiating w.r.t we get;
dy/dx = d(x⁵)/dx
y’ = 5x5-1 = 5x4
Therefore, d(x5)/dx = 5x4=20

FAQs on Differential Questions

Q: What do you mean by differentiation?

A: Differentiation permits us to discover paces of progress. For instance, it permits us to discover the pace of progress of speed as for time (which is quickening). It additionally permits us to discover the pace of progress of x as for y, which on a chart of y against x is the angle of the bend.

Q: Differentiate ln(10).

A: Since ln(10) is a constant, the derivative of ln(10) with respect to the variable “x” is 0.

Q: How can we differentiate log?

A: You can really utilize the subordinate of ln ⁡ ( x ) \ln(x) ln(x)natural log, left bracket, x, right enclosure (alongside the consistent various principle) to get the overall subsidiary of log ⁡ b ( x ) \log_b(x) logb(x)log, start base, b, end base, left bracket, x, right bracket.

Q: What is an equation to solve?

A: The equation is a mathematical expression containing the same symbol as. Equations often contain algebra. Algebra is used in mathematics when you don't know the right number in the calculation

Q: What are the types of equations?

A; There are three main forms of linear equations: point-slope shapes, standard forms, and slope-intercept forms