Bank Guides Notes Mocks Board Questions / Maharashtra State Board Class 11 / Mathematics Differentiation 87 Maharashtra State Board Class 11 Mathematics practice questions with answers and worked solutions.
Difficulty: 8 easy · 53 moderate · 26 hard. Most-asked subtopics: Rules of Differentiation (26), Derivative by the Method of First Principle (20), Relationship Between Differentiability and Continuity (13). Updated 16 August 2026. Balbharati textbook · 9.1.4a SolvedEx.1 Hard Add #1 · Derivatives of Standard Functions
Find the derivative of x n x^{n} x n w. r. t. x x x for n ∈ N n \in N n ∈ N . Balbharati textbook · 9.1.4a SolvedEx.2 Moderate Add #2 · Derivatives of Standard Functions
Find derivative of sin x \sin x sin x w. r. t. x x x . Balbharati textbook · 9.1.4a SolvedEx.3 Moderate Add #3 · Derivatives of Standard Functions
Find the derivative of tan x \tan x tan x w. r. t. x x x . Balbharati textbook · 9.1.4a SolvedEx.4 Hard Add #4 · Derivatives of Standard Functions
Find the derivative of sec x \sec x sec x w. r. t. x x x . Balbharati textbook · 9.1.4a SolvedEx.5 Moderate Add #5 · Derivatives of Standard Functions
Find the derivative of log x \log x log x w. r. t. x x x . ( x > 0 ) (x > 0) ( x > 0 ) Done a few? Here’s your next move
Balbharati textbook · 9.1.4a SolvedEx.6 Moderate Add #6 · Derivatives of Standard Functions
Find then derivative of a x a^{x} a x w. r. t. x x x . ( a > 0 ) (a > 0) ( a > 0 ) Find the derivatives of the following from the definition,
Balbharati textbook · 9.1.4 SolvedEx.1(i) Moderate Add #7 · Derivative by the Method of First Principle
Balbharati textbook · 9.1.4 SolvedEx.1(ii) Hard Add #8 · Derivative by the Method of First Principle
cos ( 2 x + 3 ) \cos(2x+3) cos ( 2 x + 3 ) Balbharati textbook · 9.1.4 SolvedEx.1(iii) Moderate Add #9 · Derivative by the Method of First Principle
Balbharati textbook · 9.1.4 SolvedEx.1(iv) Hard Add #10 · Derivative by the Method of First Principle
log ( 3 x − 2 ) \log(3x-2) log ( 3 x − 2 ) Balbharati textbook · 9.1.4 SolvedEx.2 Moderate Add #11 · Derivative by the Method of First Principle
Find the derivative of f ( x ) = sin x f(x) = \sin x f ( x ) = sin x , at x = π x = \pi x = π Balbharati textbook · 9.1.4 SolvedEx.3 Easy Add #12 · Derivative by the Method of First Principle
Find the derivative of x 2 + x + 2 x^{2}+x+2 x 2 + x + 2 , at x = − 3 x = -3 x = − 3 Balbharati textbook · 9.1.5 SolvedEx.1 Moderate Add #13 · Definition of the Derivative and Differentiability
Test whether the function f ( x ) = ( 3 x − 2 ) 2 5 f(x)=(3x-2)^{\frac{2}{5}} f ( x ) = ( 3 x − 2 ) 5 2 is differentiable at x = 2 3 x=\frac{2}{3} x = 3 2 . Balbharati textbook · 9.1.5 SolvedEx.2 Moderate Add #14 · Definition of the Derivative and Differentiability
Examine the differentiability of f ( x ) = ( x − 2 ) ∣ x − 2 ∣ f(x)=(x-2)|x-2| f ( x ) = ( x − 2 ) ∣ x − 2∣ at x = 2 x=2 x = 2 . Balbharati textbook · 9.1.5 SolvedEx.3 Hard Add #15 · Relationship Between Differentiability and Continuity
Show the function f ( x ) f(x) f ( x ) is continuous at x = 3 x=3 x = 3 , but not differentiable at x = 3 x=3 x = 3 . if
f ( x ) = 2 x + 1 f(x)=2x+1 f ( x ) = 2 x + 1 for x ≤ 3 x\le 3 x ≤ 3
= 16 − x 2 \quad\quad\;\; =16-x^2 = 16 − x 2 for x > 3 x>3 x > 3 . Balbharati textbook · 9.1.5 SolvedEx.4 Easy Add #16 · Definition of the Derivative and Differentiability
Show that the function f ( x ) f(x) f ( x ) is differentiable at x = − 3 x=-3 x = − 3 where, f ( x ) = x 2 + 2 f(x)=x^2+2 f ( x ) = x 2 + 2 . Find the derivatives of the following w. r. t. x x x by using method of first principle. Balbharati textbook · Ex 9.1 Q1(a) Easy Add #17 · Derivative by the Method of First Principle
Balbharati textbook · Ex 9.1 Q1(b) Moderate Add #18 · Derivative by the Method of First Principle
Balbharati textbook · Ex 9.1 Q1(c) Moderate Add #19 · Derivative by the Method of First Principle
Balbharati textbook · Ex 9.1 Q1(d) Moderate Add #20 · Derivative by the Method of First Principle
Balbharati textbook · Ex 9.1 Q1(e) Moderate Add #21 · Derivative by the Method of First Principle
log ( 2 x + 5 ) \log(2x+5) log ( 2 x + 5 ) Balbharati textbook · Ex 9.1 Q1(f) Hard Add #22 · Derivative by the Method of First Principle
tan ( 2 x + 3 ) \tan(2x+3) tan ( 2 x + 3 ) Balbharati textbook · Ex 9.1 Q1(g) Hard Add #23 · Derivative by the Method of First Principle
sec ( 5 x − 2 ) \sec(5x-2) sec ( 5 x − 2 ) Balbharati textbook · Ex 9.1 Q1(h) Moderate Add #24 · Derivative by the Method of First Principle
Find the derivatives of the following w. r. t. x x x . at the points indicated against them by using method of first principle Balbharati textbook · Ex 9.1 Q2(a) Moderate Add #25 · Derivative by the Method of First Principle
2 x + 5 \sqrt{2x+5} 2 x + 5 at x = 2 x=2 x = 2 Balbharati textbook · Ex 9.1 Q2(b) Moderate Add #26 · Derivative by the Method of First Principle
tan x \tan x tan x at x = π / 4 x=\pi/4 x = π /4 Balbharati textbook · Ex 9.1 Q2(c) Moderate Add #27 · Derivative by the Method of First Principle
2 3 x + 1 2^{3x+1} 2 3 x + 1 at x = 2 x=2 x = 2 Balbharati textbook · Ex 9.1 Q2(d) Moderate Add #28 · Derivative by the Method of First Principle
log ( 2 x + 1 ) \log(2x+1) log ( 2 x + 1 ) at x = 2 x=2 x = 2 Balbharati textbook · Ex 9.1 Q2(e) Moderate Add #29 · Derivative by the Method of First Principle
e 3 x − 4 e^{3x-4} e 3 x − 4 at x = 2 x=2 x = 2 Balbharati textbook · Ex 9.1 Q2(f) Moderate Add #30 · Derivative by the Method of First Principle
cos x \cos x cos x at x = 5 π 4 x=\frac{5\pi}{4} x = 4 5 π Balbharati textbook · Ex 9.1 Q3 Moderate Add #31 · Definition of the Derivative and Differentiability
Show that the function f f f is not differentiable at x = − 3 x=-3 x = − 3 ,
where f ( x ) = x 2 + 2 f(x)=x^2+2 f ( x ) = x 2 + 2 for x < − 3 x<-3 x < − 3
= 2 − 3 x \quad\quad\quad\;\; =2-3x = 2 − 3 x for x ≥ − 3 x\ge -3 x ≥ − 3 Balbharati textbook · Ex 9.1 Q4 Easy Add #32 · Relationship Between Differentiability and Continuity
Show that f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 is continuous and differentiable at x = 0 x=0 x = 0 . Discuss the continuity and differentiability of
Balbharati textbook · Ex 9.1 Q5(i) Moderate Add #33 · Relationship Between Differentiability and Continuity
f ( x ) = x ∣ x ∣ f(x)=x|x| f ( x ) = x ∣ x ∣ at x = 0 x=0 x = 0 Balbharati textbook · Ex 9.1 Q5(ii) Moderate Add #34 · Relationship Between Differentiability and Continuity
f ( x ) = ( 2 x + 3 ) ∣ 2 x + 3 ∣ f(x)=(2x+3)|2x+3| f ( x ) = ( 2 x + 3 ) ∣2 x + 3∣ at x = − 3 / 2 x=-3/2 x = − 3/2 Balbharati textbook · Ex 9.1 Q6 Hard Add #35 · Relationship Between Differentiability and Continuity
Discuss the continuity and differentiability of f ( x ) f(x) f ( x ) at x = 2 x=2 x = 2
f ( x ) = [ x ] f(x)=[x] f ( x ) = [ x ] if x ∈ [ 0 , 4 ) x\in[0,4) x ∈ [ 0 , 4 ) . [where [ ∗ ] [*] [ ∗ ] is a greatest integer ( floor ) function] Balbharati textbook · Ex 9.1 Q7 Moderate Add #36 · Relationship Between Differentiability and Continuity
Test the continuity and differentiability of
f ( x ) = 3 x + 2 f(x)=3x+2 f ( x ) = 3 x + 2 if x > 2 x>2 x > 2
= 12 − x 2 \quad\quad\;\; =12-x^2 = 12 − x 2 if x ≤ 2 x\le 2 x ≤ 2 at x = 2 x=2 x = 2 . Balbharati textbook · Ex 9.1 Q8 Hard Add #37 · Relationship Between Differentiability and Continuity
If f ( x ) = sin x − cos x f(x)=\sin x-\cos x f ( x ) = sin x − cos x if x ≤ π / 2 x\le \pi/2 x ≤ π /2
= 2 x − π + 1 \quad\quad\;\; =2x-\pi+1 = 2 x − π + 1 if x > π / 2 x>\pi/2 x > π /2 . Test the continuity and differentiability of f f f at x = π / 2 x=\pi/2 x = π /2 Balbharati textbook · Ex 9.1 Q9 Hard Add #38 · Relationship Between Differentiability and Continuity
Examine the function
f ( x ) = x 2 cos ( 1 x ) f(x)=x^2\cos\left(\frac{1}{x}\right) f ( x ) = x 2 cos ( x 1 ) , for x ≠ 0 x\ne 0 x = 0
= 0 \quad\quad\;\; =0 = 0 , for x = 0 x=0 x = 0
for continuity and differentiability at x = 0 x=0 x = 0 . Find the derivatives of the following functions
Balbharati textbook · 9.2.4 SolvedEx.1(1) Easy Add #39 · Rules of Differentiation
y = x 3 2 + log x − cos x y=x^{\frac{3}{2}}+\log x-\cos x y = x 2 3 + log x − cos x Balbharati textbook · 9.2.4 SolvedEx.1(2) Moderate Add #40 · Rules of Differentiation
f ( x ) = x 5 cosec x + x tan x f(x)=x^5\operatorname{cosec}x+\sqrt{x}\tan x f ( x ) = x 5 cosec x + x tan x Balbharati textbook · 9.2.4 SolvedEx.1(3) Moderate Add #41 · Rules of Differentiation
y = e x − 5 e x + 5 y=\frac{e^x-5}{e^x+5} y = e x + 5 e x − 5 Balbharati textbook · 9.2.4 SolvedEx.1(4) Hard Add #42 · Rules of Differentiation
y = x sin x x + sin x y=\frac{x\sin x}{x+\sin x} y = x + s i n x x s i n x Balbharati textbook · 9.2.4 SolvedEx.2 Hard Add #43 · Rules of Differentiation
If f ( x ) = p tan x + q sin x + r f(x)=p\tan x+q\sin x+r f ( x ) = p tan x + q sin x + r , f ( 0 ) = − 4 3 f(0)=-4\sqrt{3} f ( 0 ) = − 4 3 , f ( π 3 ) = − 7 3 f\left(\frac{\pi}{3}\right)=-7\sqrt{3} f ( 3 π ) = − 7 3 , f ′ ( π 3 ) = 3 f'\left(\frac{\pi}{3}\right)=3 f ′ ( 3 π ) = 3 then find p p p , q q q and r r r . Balbharati textbook · Ex 9.2 I Q1 Easy Add #44 · Derivatives of Standard Functions
Differentiate the following w.r.t. x x x : y = x 4 3 + e x − sin x y=x^{\frac{4}{3}}+e^{x}-\sin x y = x 3 4 + e x − sin x Balbharati textbook · Ex 9.2 I Q2 Easy Add #45 · Derivatives of Standard Functions
Differentiate the following w.r.t. x x x : y = x + tan x − x 3 y=\sqrt{x}+\tan x-x^{3} y = x + tan x − x 3 Balbharati textbook · Ex 9.2 I Q3 Moderate Add #46 · Derivatives of Standard Functions
Differentiate the following w.r.t. x x x : y = log x − cosec x + 5 x − 3 x 3 2 y=\log x-\operatorname{cosec} x+5^{x}-\frac{3}{x^{\frac{3}{2}}} y = log x − cosec x + 5 x − x 2 3 3 Balbharati textbook · Ex 9.2 I Q4 Moderate Add #47 · Derivatives of Standard Functions
Differentiate the following w.r.t. x x x : y = x 7 3 + 5 x 4 5 − 5 x 2 5 y=x^{\frac{7}{3}}+5x^{\frac{4}{5}}-\frac{5}{x^{\frac{2}{5}}} y = x 3 7 + 5 x 5 4 − x 5 2 5 Balbharati textbook · Ex 9.2 I Q5 Moderate Add #48 · Derivatives of Standard Functions
Differentiate the following w.r.t. x x x : y = 7 x + x 7 − 2 3 x x − log x + 7 7 y=7^{x}+x^{7}-\frac{2}{3}x\sqrt{x}-\log x+7^{7} y = 7 x + x 7 − 3 2 x x − log x + 7 7 Balbharati textbook · Ex 9.2 I Q6 Moderate Add #49 · Derivatives of Standard Functions
Differentiate the following w.r.t. x x x : y = 3 cot x − 5 e x + 3 log x − 4 x 3 4 y=3\cot x-5e^{x}+3\log x-\frac{4}{x^{\frac{3}{4}}} y = 3 cot x − 5 e x + 3 log x − x 4 3 4 Balbharati textbook · Ex 9.2 II Q1 Moderate Add #50 · Rules of Differentiation
Differentiate the following w.r.t. x x x : y = x 5 tan x y=x^{5}\tan x y = x 5 tan x Showing the first 50 of 87.
See all 87 questions Studying this with friends? Send them this chapter's questions. It's free for them too.
More chapters in this subject 