NCERT Solutions for Class 11 Maths – Chapter 12 Limits and Derivatives Exercise 12.2

NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 – Limits and Derivatives covers the concept of derivatives of functions using the first principle (limit definition) and standard differentiation rules. This exercise includes problems on finding derivatives of algebraic, polynomial, and trigonometric functions.

Prepared according to the latest CBSE Class 11 Maths syllabus 2025-26, Exercise 12.2 helps students understand how to apply the first principle of differentiation, the product rule (Leibnitz rule), and the quotient rule. Mastering these techniques is essential for Class 12 Maths and competitive exams like JEE.

NCERT Solutions for Class 11 Maths – Chapter 12 Limits and Derivatives Exercise 12.2

NCERT Solutions for Class 11 Maths – Chapter 12 Limits and Derivatives Exercise 12.2

The solutions are explained in a clear, step-by-step format to help students build a strong foundation in differential calculus.


Q1. Find the derivative of x² − 2 at x = 10.

Ans: Let f(x) = x² − 2.

f'(10) = lim(h→0) [f(10+h) − f(10)] / h

= lim(h→0) [(10+h)² − 2 − (100 − 2)] / h

= lim(h→0) [100 + 20h + h² − 2 − 100 + 2] / h

= lim(h→0) (20h + h²) / h = lim(h→0) (20 + h) = 20


Q2. Find the derivative of x at x = 1.

Ans: Let f(x) = x.

f'(1) = lim(h→0) [(1+h) − 1] / h = lim(h→0) h/h = 1


Q3. Find the derivative of 99x at x = 100.

Ans: Let f(x) = 99x.

f'(100) = lim(h→0) [99(100+h) − 99(100)] / h = lim(h→0) 99h/h = 99


Q4. Find the derivative of the following functions from first principles:

(i) x³ − 27

Ans: Let f(x) = x³ − 27.

f'(x) = lim(h→0) [(x+h)³ − 27 − (x³ − 27)] / h

= lim(h→0) [3x²h + 3xh² + h³] / h = lim(h→0) (3x² + 3xh + h²) = 3x²


(ii) (x−1)(x−2)

Ans: f(x) = x² − 3x + 2

f'(x) = lim(h→0) [(x+h)² − 3(x+h) + 2 − (x² − 3x + 2)] / h

= lim(h→0) (2xh + h² − 3h) / h = lim(h→0) (2x + h − 3) = 2x − 3


(iii) 1/x²

Ans: f(x) = 1/x²

f'(x) = lim(h→0) [1/(x+h)² − 1/x²] / h

= lim(h→0) [x² − (x+h)²] / [h · x²(x+h)²]

= lim(h→0) (−2x − h) / [x²(x+h)²] = −2x/x⁴ = −2/x³


(iv) (x+1)/(x−1)

Ans: f'(x) = lim(h→0) [(x+h+1)/(x+h−1) − (x+1)/(x−1)] / h

= lim(h→0) [(x+h+1)(x−1) − (x+1)(x+h−1)] / [h(x−1)(x+h−1)]

= lim(h→0) −2h / [h(x−1)(x+h−1)] = −2/(x−1)²


Q5. For F(x) = x¹⁰⁰/100 + x⁹⁹/99 + ... + x²/2 + x + 1, prove that f'(1) = 100·f'(0).

Ans: Differentiating: f'(x) = x⁹⁹ + x⁹⁸ + ... + x + 1

At x = 0: f'(0) = 1

At x = 1: f'(1) = 1 + 1 + 1 + ... (100 terms) = 100

Since 100 = 100 × 1 = 100 × f'(0), f'(1) = 100·f'(0) (Proved)


Q6. Find the derivative of xⁿ + axⁿ⁻¹ + a²xⁿ⁻² + ... + aⁿ⁻¹x + aⁿ.

Ans: Using d/dx(xⁿ) = nxⁿ⁻¹:

f'(x) = nxⁿ⁻¹ + a(n−1)xⁿ⁻² + a²(n−2)xⁿ⁻³ + ... + aⁿ⁻¹


Q7. For constants a and b, find the derivative of:

(i) (x−a)(x−b)

Ans: f(x) = x² − (a+b)x + ab

f'(x) = 2x − a − b


(ii) (ax² + b)²

Ans: f(x) = a²x⁴ + 2abx² + b²

f'(x) = 4a²x³ + 4abx = 4ax(ax² + b)


(iii) (x−a)/(x−b)

Ans: By quotient rule:

f'(x) = [(x−b)(1) − (x−a)(1)] / (x−b)²

= (x − b − x + a) / (x−b)² = (a−b)/(x−b)²


Q8. Find the derivative of (xⁿ − aⁿ)/(x − a) for some constant a.

Ans: By quotient rule:

f'(x) = [(x−a)·nxⁿ⁻¹ − (xⁿ − aⁿ)·1] / (x−a)²

f'(x) = [nxⁿ − anxⁿ⁻¹ − xⁿ + aⁿ] / (x−a)²


Q9. Find the derivative of:

(i) 2x − 3/4

Ans: f'(x) = 2 − 0 = 2


(ii) (5x³ + 3x − 1)(x − 1)

Ans: By Leibnitz product rule:

f'(x) = (5x³ + 3x − 1)(1) + (x−1)(15x² + 3)

= 5x³ + 3x − 1 + 15x³ + 3x − 15x² − 3

f'(x) = 20x³ − 15x² + 6x − 4


(iii) x⁻³(5 + 3x)

Ans: By Leibnitz product rule:

f'(x) = x⁻³(3) + (5 + 3x)(−3x⁻⁴)

= 3x⁻³ − 15x⁻⁴ − 9x⁻³ = −6x⁻³ − 15x⁻⁴

f'(x) = −6/x³ − 15/x⁴


(iv) x⁵(3 − 6x⁻⁹)

Ans: By Leibnitz product rule:

f'(x) = x⁵(54x⁻¹⁰) + (3 − 6x⁻⁹)(5x⁴)

= 54x⁻⁵ + 15x⁴ − 30x⁻⁵ = 24x⁻⁵ + 15x⁴

f'(x) = 15x⁴ + 24/x⁵


(v) x⁻⁴(3 − 4x⁻⁵)

Ans: By Leibnitz product rule:

f'(x) = x⁻⁴(20x⁻⁶) + (3 − 4x⁻⁵)(−4x⁻⁵)

= 20x⁻¹⁰ − 12x⁻⁵ + 16x⁻¹⁰ = 36x⁻¹⁰ − 12x⁻⁵

f'(x) = 36/x¹⁰ − 12/x⁵


(vi) 2/(x+1) − x²/(3x−1)

Ans: By quotient rule on each term:

d/dx[2/(x+1)] = −2/(x+1)²

d/dx[x²/(3x−1)] = [(3x−1)(2x) − x²(3)] / (3x−1)² = x(3x−2)/(3x−1)²

f'(x) = −2/(x+1)² − x(3x−2)/(3x−1)²


Q10. Find the derivative of cos x from first principle.

Ans: f'(x) = lim(h→0) [cos(x+h) − cos x] / h

= lim(h→0) [cos x·cos h − sin x·sin h − cos x] / h

= −cos x · lim(h→0)(1−cos h)/h − sin x · lim(h→0)(sin h/h)

= −cos x · 0 − sin x · 1

f'(x) = −sin x


Q11. Find the derivative of the following functions:

(i) sin x · cos x

Ans: f'(x) = lim(h→0) [sin(x+h)cos(x+h) − sin x·cos x] / h

= lim(h→0) [sin 2(x+h) − sin 2x] / (2h) = lim(h→0) cos(2x+h)·(sin h/h)

f'(x) = cos 2x


(ii) sec x

Ans: From first principle:

f'(x) = lim(h→0) [sec(x+h) − sec x] / h = (1/cos x)·(sin x/cos x)

f'(x) = sec x · tan x


(iii) 5 sec x + 4 cos x

Ans: f'(x) = 5·sec x·tan x + 4·(−sin x)

f'(x) = 5 sec x·tan x − 4 sin x


(iv) cosec x

Ans: From first principle:

f'(x) = −cosec x · cot x


(v) 3 cot x + 5 cosec x

Ans: Using d/dx(cot x) = −cosec²x and d/dx(cosec x) = −cosec x·cot x:

f'(x) = −3 cosec²x − 5 cosec x·cot x


(vi) 5 sin x − 6 cos x + 7

Ans: f'(x) = 5·cos x − 6·(−sin x) + 0

f'(x) = 5 cos x + 6 sin x


(vii) 2 tan x − 7 sec x

Ans: Using d/dx(tan x) = sec²x and d/dx(sec x) = sec x·tan x:

f'(x) = 2 sec²x − 7 sec x·tan x


FAQs – Class 11 Maths Chapter 12 Exercise 12.2 Limits and Derivatives

Q1. What is the first principle of differentiation? The derivative of f(x) from first principle is: f'(x) = lim(h→0) [f(x+h) − f(x)] / h. This is the fundamental definition of a derivative and is used to derive all standard differentiation formulas.

Q2. What is the Leibnitz product rule? If f(x) = u(x)·v(x), then f'(x) = u'(x)·v(x) + u(x)·v'(x). This rule is used when differentiating a product of two functions.

Q3. What is the quotient rule? If f(x) = u(x)/v(x), then f'(x) = [u'(x)·v(x) − u(x)·v'(x)] / [v(x)]². This rule is applied when differentiating a ratio of two functions.

Q4. What are the standard derivatives of trigonometric functions? d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec²x, d/dx(sec x) = sec x·tan x, d/dx(cosec x) = −cosec x·cot x, d/dx(cot x) = −cosec²x.

Q5. Why is Exercise 12.2 important for board exams? Derivatives are one of the most important topics in Class 11 and Class 12 Maths. Exercise 12.2 builds the foundation for all differentiation techniques. Questions from this exercise are frequently asked in CBSE board exams and form the basis for integral calculus and applications of derivatives.

Q6. How can students prepare effectively for Exercise 12.2? Students should memorise the first principle definition, standard derivative formulas, and the product and quotient rules. Practising all questions — especially the trigonometric derivatives and multi-step problems — will ensure thorough preparation for exams.