OP Malhotra Class 11 Maths Solutions Chapter 19 Differentiation Ex 19(d)

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S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Ex 19(d)

Question 1.
sin 5x
Solution:
Let y – sin 5x ; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sin 5x = cos 5x\(\frac{d}{d x}\)(5x) = 5 cos 5x

Question 2.
cos 8x
Solution:
Let y = cos 8x ; Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (cos 8x) = -sin8x\(\frac{d}{d x}\) (8x) = – 8 sin x

Question 3.
sin (5x + 9)
Solution:
Let y = sin (5x + 9); Diff. both sides w.r.t. x,
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sin (5x + 9) = cos(5x + 9)\(\frac{d}{d x}\)(5x + 9) = cos (5x + 9) (5.1 + 0) = 5 cos (5x + 9)

Question 4.
cos (2x – 3)
Solution:
Let y = cos (2x – 3) ; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)cos(2x – 3) = -sin(2x – 3)\(\frac{d}{d x}\)(2x – 3) = – 2 sin (2x – 3)

Question 5.
tan 7x
Solution:
Let y = tan 7x ; Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) tan 7x = sec² 7x\(\frac{d}{d x}\) (7x) = 7 sec² 7x

Question 6.
cot nx
Solution:
Let y = cot nx ; Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) cot nx = – cosec² nx\(\frac{d}{d x}\) (nx) = – n cosec² nx

Question 7.
tan (6x + 11)
Solution:
Let y = tan (6x + 11); Diff. both sides w.r.t. x
\(\frac{d y}{d x}\) = sec²(6x +11) \(\frac{d}{d x}\)(6x + 11) = sec² (6x + 11) (6.1 + 0) = 6 sec² (6x + 11)

Question 8.
sin \(\frac{x}{3}\)
Solution:
Let y = sin\(\frac{x}{3}\); diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = cos\(\frac{x}{3}\) \(\frac{d}{d x}\) \(\frac{x}{3}\) = \(\frac{1}{3}\) cos \(\frac{x}{3}\)

Question 9.
sec mx
Solution:
Let y = sec mx ; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (sec mx) = sec mx tan mx \(\frac{d}{d x}\) (mx) = m sec mx tan mx

Question 10.
Solution:
Let y = sec \(\left(\frac{x}{2}-1\right)\); Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sec\(\left(\frac{x}{2}-1\right)\) = sec\(\left(\frac{x}{2}-1\right)\) tan \(\left(\frac{x}{2}-1\right)\) \(\frac{d}{d x}\) \(\left(\frac{x}{2}-1\right)\) = \(\frac{1}{2}\) sec \(\left(\frac{x}{2}-1\right)\) tan \(\left(\frac{x}{2}-1\right)\)

Question 11.
cosec \(\frac{2}{3}\)x
Solution:
Let y = cosec \(\frac{2}{3}\)x ; Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = – cot\(\frac{2}{3}\) x cosec \(\frac{2}{3}\)x \(\frac{d}{dx}\) \(\left(\frac{2}{3} x\right)\) = –\(\frac{2}{3}\) cot \(\frac{2}{3}\) x cosec \(\frac{2x}{3}\)

Question 12.
x sin x
Solution:
Let y = x sin x; Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x sin x) = x\(\frac{d}{d x}\) sin x + sin x \(\frac{d}{d x}\)(x)
[∵ \(\frac{d}{d x}\) (uv) = u\(\frac{dv}{d x}\) + v \(\frac{du}{d x}\)]
= x cos x + sin x . 1 = x cos x + sin x

Question 13.
x² cos 5x
Solution:
Let y = x² cos 5x; Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = x² \(\frac{d}{d x}\) cos 5x + cos 5x \(\frac{d}{d x}\)x²
[∵ \(\frac{d}{d x}\) (uv) = u\(\frac{dv}{d x}\) + v \(\frac{du}{d x}\)]
= x² (-sin 5x) \(\frac{d}{d x}\) (5x) + cos 5x . 2x = – 5x² sin 5x + 2x cos 5x

Question 14.
\(\sqrt{x}\) cosec (5x + 7)
Solution:
Let y = \(\sqrt{x}\) cosec (5x + 7); Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\sqrt{x}\) \(\frac{d}{d x}\)cosec (5x + 7) + cosec (5x + 7) \(\frac{d}{d x}\) = \(\sqrt{x}\)
= \(\sqrt{x}\) {- cot (5x + 7) cosec (5x + 7)} \(\frac{d}{d x}\) (5x + 7) + cosec (5x + 7) \(\frac{1}{2} x^{\frac{1}{2}-1}\)
= -5\(\sqrt{x}\) cot (5x + 7) cosec (5x + 7) + \(\frac{1}{2 \sqrt{x}}\) cosec (5x + 7)

Question 15.
\(\frac{\sin 3 x}{x-6}\)
Solution:
Let y = \(\frac{\sin 3 x}{x-6}\); Diff. both sides w.r.t. x, we have

Question 16.
\(\frac{\cos x}{5 x}\)
Solution:

Question 17.
\(\frac{\tan x}{2 x+3}\)
Solution:
Let y = \(\frac{\tan x}{2 x+3}\); Diff. both sides w.r.t. x, we have

Question 18.
\(\frac{\sec (a x-b)}{x^2-2}\)
Solution:
Let y = \(\frac{\sec (a x-b)}{x^2-2}\); diff. both sides w.r.t. x, we have

Question 19.
sin 2x
Solution:
Let y = sin 2x; Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) sin 2x = cos 2x \(\frac{d}{d x}\) (2x) = 2 cos 2x

Question 20.
cos 3x
Solution:
Let y = cos 3x; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) (cos 3x) = – sin 3x\(\frac{d }{d x}\) (3x) = – 3 sin 3x

Question 21.
tan 2x
Solution:
Let y = tan 2x; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) (tan 2x) = sec² 2x\(\frac{d }{d x}\) (2x) = 2 sec² 2x

Question 22.
sin\(\frac{x}{2}\)
Solution:
Let y = sin\(\frac{x}{2}\); diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) sin \(\frac{x}{2}\) = cos\(\frac{x}{2}\) \(\frac{d}{dx}\) \(\frac{x}{2}\) = \(\frac{1}{2}\) cos \(\frac{x}{2}\)

Question 23.
sec ax
Solution:
Let y = sec ax ; diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) (sec ax) = sec ax tan ax \(\frac{d }{d x}\)(ax) = a sec ax tan ax

Question 24.
sec (px + q)
Solution:
Let y = sec (px + q); diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) = sec(px + q) = sec (px + q) tan (px + q) \(\frac{d }{d x}\) (px + q) = p sec (px + q) tan (px + q)

Question 25.
tan (4x – 7)
Solution:
Let y = tan (4x – 7) ; diff. both sides w.r.t. x,
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) tan(4x – 7) = sec²(4x – 7) \(\frac{d }{d x}\)(4x – 7)
= sec² (4x – 7) (4 × 1 – 0) = 4 sec² (4x – 7)