OP Malhotra Class 11 Maths Solutions Chapter 19 Differentiation Chapter Test

Continuous practice using ISC OP Malhotra Solutions Class 11 Chapter 19 Differentiation Chapter Test can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Chapter Test

Question 1.
Find from first principles the differential coefficient of 2x² + 3x.
Solution:
Let y =f (x) = 2x² + 3x
∴ f(x + δx) = 2 (x + δx)² + 3 (x + δx)
Then by first principle, we have

Question 2.
Find from first principles the differential coefficient of sin 2x.
Solution:
Let y = f(x) = sin 2x
∴ f(x + δx) = sin 2 (x + δx)
Then by first principle, we have

Question 3.
f(x) = \(\sqrt{3 x+4}\), x > – 1
Solution:
Given f(x) = \(\sqrt{3 x+4}\); Diff. both sides w.r.t. x, we have
\(\frac{d}{d x}\) f(x) = f ‘ (x) = \(\frac{d{d x}\) (3x + 4)^1/2

Question 4.
f(x) = \(\sqrt{4-x}\), x < 4
Solution:
Given f(x) = \(\sqrt{4-x}\); Diff. both sides w.r.t. x, we have

Question 5.
f(x) = \(\frac{3 x+4}{4 x+3}\left(x \neq \frac{-3}{4}\right)\)
Solution:
Given f(x) = \(\frac{3 x+4}{4 x+3}\); Diff. both sides w.r.t. x, we have

Question 6.
f(x) = \(\sqrt{x^2+1}\)
Solution:
Given f(x) = \(\sqrt{x^2+1}\); Diff. both sides w.r.t. x, we have

Question 7.
(2x + 3) (x² – x + 2)
Solution:
Let Y = (2x + 3) (x² – x + 2); Diff. both sides w.r.t. x, we have
\(\frac{dy}{d x}\) = (2x + 3) \(\frac{d}{d x}\) (x² – x + 2) + (x² – x + 2) \(\frac{d}{d x}\) (2x + 3)
[∵\(\frac{d}{d x}\) (uv) = \(\frac{udv}{d x}\) + v\(\frac{du}{d x}\)]
= (2x + 3) (2x – 1) + (x² – x + 2) . 2 = 4x² + 4x – 3 + 2x² – 2x + 4 = 6x² + 2x + 1

Question 8.
tan (5x + 7)
Solution:
Let y = tan (5x + 7); Diff. both sides w.r.t. x, we have
\(\frac{dy}{d x}\) = \(\frac{d}{d x}\) tan (5x + 7) = sec² (5x + 7)
= sec² (5x + 7) [5.1 + 0] = 5 sec² (5x + 1)

Question 9.
sin² (3x – 2)
Solution:
Let y = sin² (3x – 2); Diff. both sides w.r.t. x, we have
\(\frac{dy}{d x}\) = \(\frac{d}{d x}\) [sin (3x – 2)]² = 2 sin (3x – 2) \(\frac{d}{d x}\) sin (3x – 2)
= 2 sin (3x – 2) cos (3x – 2) . 3 = 3 sin 2 (3x – 2)
= 3 sin (6x – 4)

Question 10.
(x³ + sin x)⁵
Solution:
Let y = (x³ + sin x)⁵; Diff. both sides w.r.t. x, we have
\(\frac{dy}{d x}\) = 5 (x³ + sin x)⁴ \(\frac{d}{d x}\) (x³ + sin x) = 5 (x³ + sin x)⁴ (3x² + cos x)