ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.5

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.5

Question 1.
(i) If f(x) = \(\sqrt{4-x^2}\), x ∈ (- 2, 2), find f'(x).
(ii) If f(x) = \(\sqrt{1-x^2}\), x ∈ (0, 1), find f'(x).
Solution:
(i) Given f(x) = \(\sqrt{4-x^2}\) ; x ∈ (- 2, 2)
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{d}{d x}\) (4 – x²)^{\(\frac{1}{2}\)}
= \(\frac{1}{2}\) (4 – x²)^{\(\frac{1}{2}\)}
= \(\frac{1}{2} \frac{1}{\sqrt{4-x^2}}\) (0 – 2x)
= \(\frac{-x}{\sqrt{4-x^2}}\), x ∈ (- 2, 2).

(ii) Given f(x) = \(\sqrt{1-x^2}\) ; x ∈ (0, 1)
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{1}{2}\) (1 – x²)^{\(-\frac{1}{2}\)} \(\frac{d}{d x}\) (1 – x²)
= \(\frac{1}{2 \sqrt{1-x^2}}\) (- 2x)
= \(\frac{-x}{\sqrt{1-x^2}}\), x ∈ (0, 1).

Question 2.
(i) \(\sqrt{2x-1}\)
(ii) (3x² – 9x + 5)⁹ (NCERT)
Solution:
(i) Let y = \(\sqrt{2x-1}\) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{2}(2 x-1)^{\frac{1}{2}-1} \frac{d}{d x}(2 x-1)\)
= \(\frac{1}{2 \sqrt{2 x-1}}\) (2 – 0)
= \(\frac{1}{\sqrt{2 x-1}}\)

(ii) Let y = (3x² – 9x + 5)⁹
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 9 (3x²– 9x + 5)^9-1 \(\frac{d}{d x}\) (3x²– 9x + 5)
[∵ \(\frac{d}{d x}\) xⁿ = nx^n-1]
= 9 (3x² – 9x + 5)⁸ (6x – 9)
= 27 (3x² – 9x + 5)⁸ (2x – 3)

Question 3.
(i) \(\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^2+4}}\) (NCERT)
(ii) sin³ x + cos⁶ x (NCERRT)
Solution:
(i) Let y = \(\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^2+4}}\)
Diff. both sides w.r.t. x, we have
\(\)
= \(\frac{1}{2 \sqrt{3 x+2}}(3+0)-\frac{1}{2\left(2 x^2+4\right)^{\frac{3}{2}}}(4 x+0)\)
= \(\frac{3}{2 \sqrt{3 x+2}}-\frac{2 x}{\left(2 x^2+4\right)^{3 / 2}}\), x > – \(\frac{2}{3}\)

(ii) Let y = sin³ x + cos⁶ x
= (sin x)³ + (cos x)⁶
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 3 (sin x)² \(\frac{d}{d x}\) (sin x) + 6 (cos x)⁵ \(\frac{d}{d x}\) (cos x)
= 3 sin² x cos x – 6 cos⁵ x sin x
= 3 sin x cos x (sin x – 2 cos⁴ x)

Question 4.
(i) sin (x²) (NCERT)
(ii) cos √x (NCERT)
Solution:
(i) Let y = sin (x²) ;
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = cos x² \(\frac{d}{d x}\) (x²
= 2x cos x²

(ii) Let y = cos √x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – sin √x \(\frac{d}{d x}\) √x
= \(\frac{-\sin \sqrt{x}}{2 \sqrt{x}}\)

Question 5.
(i) sin (ax + b) (NCERT)
(ii) tan (2x + 3) (NCERT)
Solution:
(i) Let y = sin (ax + b);
Diff. w.r.t. x, we have
\(\frac{d y}{d x}\) = cos (ax + b) \(\frac{d}{d x}\) (ax + b)
= a cos (ax + b).

(ii) Let y = tan (2x + 3) ;
Diff. w.r.t. x, we have
\(\frac{d y}{d x}\) = sec² (2x + 3) \(\frac{d}{d x}\) (2x + 3)
= sec² (2x + 3) (2 + 0)
= 2 sec² (2x + 3)

Question 6.
(i) cos (sin x) (NCERT)
(ii) sin (cos (x²)) (NCERT)
Solution:
(i) Let y = cos (sin x) ;
Diff. w.r.t. x, we have
\(\frac{d y}{d x}\) = – sin (sin x) \(\frac{d}{d x}\) (sin x)
= – cos x sin (sin x)

(ii) Let y = sin (cos (x²)
Diff. w.r.t. x, we have
\(\frac{d y}{d x}\) = cos (cos (x²)) \(\frac{d}{d x}\) cos (x²)
= cos (cos (x²)) (- sin (x²)) \(\frac{d}{d x}\) x²
= – 2x cos (cos (x²)) sin (x²)

Question 7.
(i) cos (tan \(\sqrt{x+1}\)) (NCERT Exampler)
(ii) \(\sqrt{\tan \sqrt{x}}\) (NCERT Exampler)
Solution:
(i) Let y = cos (tan \(\sqrt{x+1}\)) ;
Diff. w.r.t. x, we have
\(\frac{d y}{d x}\) = – sin (tan \(\sqrt{x+1}\)) \(\frac{d}{d x}\) tan (\(\sqrt{x+1}\))
= – sin (tan \(\sqrt{x+1}\)) sec² (\(\sqrt{x+1}\)) \(\frac{d}{d x}\) \(\sqrt{x+1}\)
= – sin (tan \(\sqrt{x+1}\)) sec² (\(\sqrt{x+1}\)) \(\frac{1}{2}\) (x + 1)^{\(\frac{1}{2}\) – 1} \(\frac{d}{d x}\) (x + 1)
= – sin (tan \(\sqrt{x+1}\)) sec² (\(\sqrt{x+1}\)) \(\frac{1}{2 \sqrt{x+1}}\) (0 + 1)
= – \(\frac{1}{2 \sqrt{x+1}}\) sin (tan \(\sqrt{x+1}\)) sec² (\(\sqrt{x+1}\))

(ii) Let y = \(\sqrt{\tan \sqrt{x}}\) ;
Diff. w.r.t. x, we have
\(\frac{d y}{d x}=\frac{d}{d x}(\tan \sqrt{x})^{\frac{1}{2}}\)
= \(\frac{1}{2}(\tan \sqrt{x})^{\frac{1}{2}-1} \frac{d}{d x} \tan \sqrt{x}\)
= \(\frac{1}{2 \sqrt{\tan \sqrt{x}}} \sec ^2 \sqrt{x} \frac{d}{d x} \sqrt{x}\)
= \(\frac{\sec ^2 \sqrt{x}}{2 \sqrt{\tan \sqrt{x}}} \frac{d}{d x} x^{1 / 2}\)
= \(\frac{\sec ^2 \sqrt{x}}{2 \sqrt{\tan \sqrt{x}}} \frac{1}{2} x^{\frac{1}{2}-1}\)
= \(\frac{\sec ^2 \sqrt{x}}{4 \sqrt{x} \sqrt{\tan \sqrt{x}}}\)

Question 8.
(i) If f(x) = x – 1, find \(\frac{d}{d x}\) ((fof) (x)).
(ii) If f (x) = x + 7 and g(x) = x – 7, find \(\frac{d}{d x}\) ((fof (x)).
Solution:
(i) Given f(x) = x – 1
∴ (fof) (x) = f (f(x))
= f(x – 1)
= x – 1 – 1
= x – 2
⇒ \(\frac{d}{d x}\) (fof) (x) = \(\frac{d}{d x}\) (x – 2)
= 1 – 0 = 1

Question 9.
If f(x) = \(\frac{x^{100}}{100}+\frac{x^{99}}{99}+\frac{x^{98}}{98}\) + …………. + x + 1, show that f'(1) = 100 f'(0)
Solution:
Given f(x) = \(\frac{x^{100}}{100}+\frac{x^{99}}{99}+\frac{x^{98}}{98}\) + …………. + x + 1 …………….(1)
f'(x) = \(\frac{100 x^{99}}{100}+\frac{99 x^{98}}{99}+\frac{98 x^{97}}{98}\) + ………. + 1
[after differentiating (1) w.r.t. x, we have]
⇒ f'(x) = x⁹⁹ + x⁹⁸ + x⁹⁷ + ………. + x + 1
∴ f'(x) = 1⁹⁹ + 1⁹⁸ + 1⁹⁷ ……… + 1
= 1 + 1 + …….. + 100 times = 100 ……………(1)
and f'(0) = 0 + 0 + 0 + ………… + 0 + 1 = 1
∴ 100 f'(0) = 100 …………….(2)
From (1) and (2) ; we have
f'(1) = 100 f'(0).

Question 10.
Find the derivative of |2×2 – 3| with respect to x.
Solution:
Let y = |2×2 – 3| ;
Diff. bothsides w.r.t. x, we have
∴ \(\frac{d y}{d x}\) = [\(\frac{2 x^2-3}{\left|2 x^2-3\right|}\) \(\frac{d}{d x}\) (2x² – 3)
[∵ \(\frac{d}{d x}\) |x| = \(\frac{x}{|x|}\), x ≠ 0]
= \(\frac{2 x^2-3}{\left|2 x^2-3\right|}\) (4x – 0)
= \(\frac{4 x\left(2 x^2-3\right)}{\left|2 x^2-3\right|}\).

Question 11.
If f(x) = |cos x|, find f'(\(\frac{3 \pi}{4}\)).
Solution:
Given f(x) = | cos x |
Diff. bothsides w.r.t. x, we have

Differentiating the following (12 to 15) functions w.r.t x :

Question 12.
(i) \(\sqrt{\frac{a^2-x^2}{a^2+x^2}}\)
(ii) sin √x + cos² √x. (NCERT Exampler)
Solution:
(i) Let y = \(\sqrt{\frac{a^2-x^2}{a^2+x^2}}\)
Diff. bothsides w.r.t. x, we have

(ii) Let y = sin √x + cos √x ;
Diff. bothsides w.r.t. x, we have
\(\frac{d y}{d x}\) = cos √x \(\frac{d}{d x}\) √x + \(\frac{d}{d x}\) (cos √x)²
= cos √x . \(\frac{1}{2}\) x^{\(\frac{1}{2}\) – 1} + 2 (cos √x)^2-1 \(\frac{d}{d x}\) cos √x
= \(\frac{\cos \sqrt{x}}{2 \sqrt{x}}\) + 2 cos √x (- sin √x) \(\frac{d}{d x}\) x^1/2
= \(\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{2 \cos \sqrt{x} \sin \sqrt{x}}{2 \sqrt{x}}\)
= \(\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin (2 \sqrt{x})}{2 \sqrt{x}}\)
[∵ sin 2θ = 2 sin θ cos θ]
= \(\frac{\cos \sqrt{x}-\sin (2 \sqrt{x})}{2 \sqrt{x}}\).

Question 13.
(i) \(\frac{\sin (a x+b)}{\cos (c x+d)}\) (NCERT)
(ii) \(\frac{\sin x+x^2}{\cot 2 x}\)
Solution:
(i) Let y = \(\frac{\sin (a x+b)}{\cos (c x+d)}\)
Diff. bothsides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\cos (c x+d) \frac{d}{d x} \sin (a x+b)-\sin (a x+b) \frac{d}{d x} \cos (c x+d)}{\cos ^2(c x+d)}\)
= \(\frac{\cos (c x+d) \cos (a x+b) \cdot a+\sin (a x+b) \sin (c x+d) \cdot c}{\cos ^2(c x+d)}\)

(ii) Let y = \(\frac{\sin x+x^2}{\cot 2 x}\)
⇒ y = sin x tan 2x + x² tan 2x
Diff. bothsides w.r.t. x, we have
\(\frac{d y}{d x}\) = sin x \(\frac{d }{d x}\) (tan 2x) + tan 2x \(\frac{d}{d x}\) (sin x) + x² \(\frac{d}{d x}\) (tan 2x) + tan 2x \(\frac{d}{d x}\) x².
= sin x . sec² 2x . 2 + tan 2x cos x + x² sec² 2x . 2 + 2x tan 2x
= 2 (sin x + x²) sec² 2x + (2x + cos x) tan 2x.

Question 14.
(i) sin^m x cosⁿ x (NCERT Exampler)
(ii) sinⁿ (ax² + bx + c) (NCERT Exampler)
Solution:
(i) Let y = sin^m x cosⁿ x ;
diff. bothsides w.r.t. x, we get
\(\frac{d y}{d x}\) = sin^m x \(\frac{d}{d x}\) (cos x)ⁿ + cosⁿ x \(\frac{d}{d x}\) (sin x)^m
= sin^m x . n (cos x)^n-1 (- sin x) + cosⁿ x . m (sin x)^m-1 . cos x
= – n sin^m+1 x cos^n-1 x + m (sin x)^m-1 cosⁿ⁺¹ x
= sin^n-1 x cos^n-1 x (m cos² x – n sin² x)

(ii) Let y = sinⁿ (ax² + bx + c)
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [sin (ax² + bx + c)]ⁿ
= n {sin (ax² + bx + c)}^n-1 \(\frac{d}{d x}\) sin (ax² + bx + c)
= n sin^n-1 (ax² + bx + c) cos (ax² + bx + c) \(\frac{d}{d x}\) (ax² + bx + c)
= n sin^n-1 (ax² + bx + c) cos (ax² + bx + c) (2ax + b)

Question 15.
(i) \(\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}\)
(ii) \(\frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}\)
Solution:
(i) Let y = \(\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}\)

(ii) Let y = \(\frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}} \times \frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}}\)
⇒ y = \(\frac{\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)^2}{x^2+1-\left(x^2-1\right)}\)
= \(\frac{1}{2}\) [x² + 1 + x² – 1 + 2 \(\sqrt{x^4-1}\)]
⇒ y = \(\frac{1}{2}\) [2x² + 2 \(\sqrt{x^4-1}\)]
= x² + \(\sqrt{x^4-1}\)
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}\) = 2x + \(\frac{1}{2}\) (x⁴ – 1)^{\(\frac{1}{2}\) – 1} \(\frac{d}{d x}\) (x⁴ – 1)
= 2x + \(\frac{1}{2 \sqrt{x^4-1}}\) (4x³)
= 2x + \(\frac{2 x^3}{\sqrt{x^4-1}}\)

Question 16.
If f(x) = \(\sqrt{\frac{\sec x-1}{\sec x+1}}\), find f'(x). Also find f’\(\frac{\pi}{3}\)).
Solution:
Given f(x) = \(\sqrt{\frac{\sec x-1}{\sec x+1}}\)
= \(\sqrt{\frac{1-\cos x}{1+\cos x}}\)
= \(\sqrt{\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}}\)
⇒ f(x) = tan \(\frac{x}{2}\) ;
Diff. bothsides w.r.t. x , we get
f'(x) = sec² \(\frac{x}{2}\) . \(\frac{1}{2}\)
at x = \(\frac{\pi}{3}\), we get
f'(\(\frac{\pi}{3}\)) = \(\sec ^2 \frac{\pi}{6} \cdot \frac{1}{2}\)
= \(\frac{1}{2}\left(\frac{2}{\sqrt{3}}\right)^2\)
= \(\frac{1}{2} \times \frac{4}{3}=\frac{2}{3}\).

Question 17.
Find the derivatives of the following functions at the inclined points:
(i) cos (2x + \(\frac{\pi}{3}\)) at x = \(\frac{\pi}{3}\)
(ii) \(\frac{1-\sin x}{1+\cos x}\) at x = \(\frac{\pi}{2}\)
Solution:
(i) Let y = cos (2x + \(\frac{\pi}{3}\)) ;
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}\) = – sin (2x + \(\frac{\pi}{3}\)) \(\frac{d}{d x}\) (2x + \(\frac{\pi}{3}\))
= – 2 sin (2x + \(\frac{\pi}{3}\))
at x = \(\frac{\pi}{3}\) ;
\(\frac{d y}{d x}\) = – 2 sin \(\left(\frac{2 \pi}{3}+\frac{\pi}{3}\right)\)
= – 2 sin π
= – 2 × 0 = 0.

(ii) Let y = \(\frac{1-\sin x}{1+\cos x}\) ;
Diff. bothsides w.r.t. x , we get

Question 18.
If y = √x + \(\frac{1}{\sqrt{x}}\), then show that 2x \(\frac{d y}{d x}\) + y = 2√x.
Solution:
Given y = √x + \(\frac{1}{\sqrt{x}}\)
Diff. (1) bothsides w.r.t. x , we get
\(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{x}}-\frac{1}{2} x^{-3 / 2}\)
= \(\frac{1}{2 \sqrt{x}}-\frac{1}{2 x^{3 / 2}}\)
⇒ 2x \(\frac{d y}{d x}\) = 2x \(\left(\frac{1}{2 \sqrt{x}}-\frac{1}{2 x^{3 / 2}}\right)\)
= √x – \(\frac{1}{\sqrt{x}}\)
⇒ 2x \(\frac{d y}{d x}\) + y = √x – \(\frac{1}{\sqrt{x}}\) + y
= √x – \(\frac{1}{\sqrt{x}}\) + √x + \(\frac{1}{\sqrt{x}}\) [using (1)]
= 2√x.

Question 19.
If y = \(\sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}}\), prove that \(\frac{d y}{d x}\) + sec² (\(\frac{\pi}{4}\) – x) = 0.
Solution:
Given y = \(\sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}}\)
= \(\sqrt{\frac{\sin ^2 x+\cos ^2 x-2 \sin x \cos x}{\sin ^2 x+\cos ^2 x+2 \sin x \cos x}}\)
= \(\sqrt{\frac{(\cos x-\sin x)^2}{(\cos x+\sin x)^2}}\)
= \(\frac{\cos x-\sin x}{\cos x+\sin x}\)
Divide Numerator and deno. by cos x ; we have
y = \(\sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}}\)
= tan (\(\frac{\pi}{4}\) – x)
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}=\sec ^2\left(\frac{\pi}{4}-x\right) \frac{d}{d x}\left(\frac{\pi}{4}-x\right)\)
\(\frac{d y}{d x}\) = sec² (\(\frac{\pi}{4}\) – x) (0 – 1)
\(\frac{d y}{d x}\) + sec² (\(\frac{\pi}{4}\) – x) = 0.

Question 20.
Differentiate | cos x | w.r.t. x. Is this function differentiable ? What can you say about the differentiability of cos |x| ?
Solution:
Let f(x) = | cos x | ;
Diff. bothsides w.r.t. x , we get
∴ f'(x) = \(\frac{\cos x}{|\cos x|}\) \(\frac{d}{d x}\) cos x
[∵ \(\frac{d}{d x}\) |x| = \(\frac{x}{|x|}\) ; x ≠ 0]
∴ f'(x) = \(\frac{-\sin x \cos x}{|\cos x|}\), provided cos x ≠ 0
= \(-\frac{\sin x \cos x}{|\cos x|}\), provided x ∈ R – {odd multiple of \(\frac{\pi}{2}\)} or
x ≠ (2n + 1) \(\frac{\pi}{2}\) ∀ n ∈ I
Since f'(x) does not exists at x = (2n + 1) \(\frac{\pi}{2}\), ∀ n ∈ I
But x = (2n + 1) \(\frac{\pi}{2}\), n ∈ I ∈ domain of cos x, which is all reals R.
Hence | cos x | is not differentiable.
Also cos |x| = \(\left\{\begin{array}{cc}
\cos x & ; \quad x \geq 0 \\
\cos (-x) & ; \quad x<0
\end{array}\right.\)
= \(\begin{cases}\cos x ; & x \geq 0 \\ \cos x ; & x<0\end{cases}\)
∴ cos |x| = cos x, which is differentiable everywhere.
Hence cos |x| is differentiable everywhere.

Question 21.
Prove that the derivative of an add function is always an even functions.
Solution:
Let f(x) be an odd function.
∴ f(- x) = – f(x)
differentiating both sides w.r.t. x,
f'(- x) \(\frac{d}{d x}\) (- x) = – \(\frac{d}{d x}\) f(x)
⇒ – f'(- x) = – f'(x)
⇒ f’ (- x) = f'(x)
Thus f'(x) is an even function.
Hence derivative of an odd function is always an even function.

Question 22.
If 2 f(x) + 3 f(- x) = x² + x + 1, find f'(1).
Solution:
Given 2 f(x) + 3 f(- x) = x² + x + 1
differentiating both sides w.r.t. x,
2f'(x) – 3 f'(x) = 2x + 1 ……………(1)
putting x = 1 on both sides of eqn. (1) ;
2f'(1) – 3f'(- 1) = 3 ………….(2)
putting x = – 1 in eqn. (2) ; we have
2f'(- 1) – 3 f'(1) = – 1 …………(3)
eqn. (2) × 2 + 3 × eqn. (3)
4f'(1) – 9 f'(1) = 6 – 3
⇒ – 5f'(1) = 3
⇒ f'(1) = – \(\frac{3}{5}\)

Exercise 5.5 (old)

Differentiate the following (1 to 7) functions :

Question 1.
(i) 5 tan x – 3 sin x + 4x^3/2
(ii) 3 cot x + 6 (1 – 2x)^5/3
Solution:
(i) Let y = 5 tan x – 3 sin x + 4x^3/2
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}=5 \frac{d}{d x}(\tan x)-3 \frac{d}{d x}(\sin x)+4 \frac{d}{d x} x^{3 / 2}\)
= 5 sec² x – 3 cos x + 4 \(\frac{3}{2}\) x^{\(\frac{3}{2}\) – 1}
= 5 sec² x – 3 cos x + 6√x

(ii) Let y = 3 cot x + 6 (1 – 2x)^5/3
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}=3 \frac{d}{d x} \cot x+6 \frac{d}{d x}(1-2 x)^{\frac{5}{3}}\)
= – 3 cosec² x + 6 × \(\frac{5}{3}\) (1 – 2x)^{\(\frac{5}{3}\) – 1} \(\frac{d}{d x}\) (1 – 2x)
= – 3 cosec² x + 10 (1 – 2x)^2/3 (0 – 2)
= – 3 cosec² x – 20 (1 – 2x)^2/3

Question 2.
(i) x³ sin x
(ii) (1 + x²) cos x
Solution:
(i) Let y = x³ sin x
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x³ sin x) + sin x \(\frac{d}{d x}\) x³
[using product rule]
= x³ cos x + sin x . 3x²
= x² (x cos x + 3 sin x)

(ii) Let y = (1 + x²) cos x
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}\) = (1 + x²) \(\frac{d}{d x}\) (cos x) + cos x \(\frac{d}{d x}\) (1 + x²)
= (1 + x²) (- sin x) + cos x (0 + 2x)
= 2x cos x – (1 + x²) sin x

Question 3.
(i) x³ cosec x
(ii) (1 – 2 tan x) (5 + 4 sin x)
Solution:
Let y = x³ cosec x
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}\) = x³ \(\frac{d}{d x}\) (cosec x) + cosec x \(\frac{d}{d x}\) x³
[using product rule]
= x³ (- cot x cosec x) + cosec x (3x²)
= x² cosec x (3 – x cot x)

(ii) Let y = (1 – 2 tan x) (5 + 4 sin x)
Diff. bothsides w.r.t. x , we get
\(\frac{d y}{d x}\) = (1 – 2 tan x) \(\frac{d}{d x}\) (5 + 4 sin x) + (5 + 4 sin x) \(\frac{d}{d x}\) (1 – 2 tan x)
[using product rule]
= (1 – 2 tan x) (4 cos x) + (5 + 4 sin x) (- 2 sec² x)

Question 4.
(i) \(\frac{x+\sin x}{x+\cos x}\)
(ii) \(\frac{\sin x+\cos x}{\sin x-\cos x}\)
Solution:
Let y = \(\frac{x+\sin x}{x+\cos x}\)
Diff. bothsides w.r.t. x , we get
∴ \(\frac{d y}{d x}\) = \(\frac{(x+\cos x) \frac{d}{d x}(x+\sin x)-(x+\sin x) \frac{d}{d x}(x+\cos x)}{(x+\cos x)^2}\)
[using quotient rule]
= \(\frac{(x+\cos x)(1+\cos x)-(x+\sin x)(1-\sin x)}{(x+\cos x)^2}\)
= \(\frac{x(\cos x+\sin x)+\cos x+\cos ^2 x-\sin x+\sin ^2 x}{(x+\cos x)^2}\)
= \(\frac{x(\cos x+\sin x)+\cos x-\sin x+1}{(x+\cos x)^2}\)

(ii) Let y = \(\frac{\sin x+\cos x}{\sin x-\cos x}\)
Diff. bothsides w.r.t. x , we get

Question 5.
(i) \(\frac{\sec x-1}{\sec x+1}\)
(ii) \(\frac{a+\sin x}{1+a \sin x}\)
Solution:
(i) Let y = \(\frac{\sec x-1}{\sec x+1}\)
= \(\frac{1-\cos x}{1+\cos x}\)
Diff. bothsides w.r.t. x , we have
\(\frac{d y}{d x}\) = \(\frac{(1+\cos x) \frac{d}{d x}(1-\cos x)-(1-\cos x) \frac{d}{d x}(1+\cos x)}{(1+\cos x)^2}\)
[using quotient rule]
= \(\frac{(1+\cos x) \sin x-(1-\cos x)(-\sin x)}{(1+\cos x)^2}\)
= \(\frac{\sin x+\cos x \sin x+\sin x-\cos x \cos x}{(1+\cos x)^2}\)
= \(\frac{2 \sin x}{(1+\cos x)^2}\)

(ii) Let y = \(\frac{a+\sin x}{1+a \sin x}\)
Diff. bothsides w.r.t. x , we have
\(\frac{d y}{d x}\) = \(\frac{(1+a \sin x) \frac{d}{d x}(a+\sin x)-(a+\sin x) \frac{d}{d x}(1+a \sin x)}{(1+a \sin x)^2}\)
[using quotient rule]
= \(\frac{(1+a \sin x)(\cos x)-(a+\sin x)(a \cos x)}{(1+a \sin x)^2}\)
∴ \(\frac{d y}{d x}=\frac{\left(1-a^2\right) \cos x}{(1+a \sin x)^2}\).

Question 6.
(i) \(\frac{x \sin x}{1+\cos x}\)
(ii) \(\frac{\sin x-x \cos x}{x \sin x+\cos x}\)
Solution:
Let y = \(\frac{x \sin x}{1+\cos x}\) ;
Diff. bothsides w.r.t. x , we have

(ii) Let y = \(\frac{\sin x-x \cos x}{x \sin x+\cos x}\) ;
Diff. bothsides w.r.t. x , we have

Question 7.
(i) \(\frac{a \cos x+b \sin x+c}{\sin x}\) + xⁿ cos x
(ii) \(\sqrt{\frac{\sec x+\tan x}{\sec x-\tan x}}\)
Solution:
(i) Let y = \(\frac{a \cos x+b \sin x+c}{\sin x}\) + xⁿ cos x
⇒ y = a cot x + b c cosec x + xⁿ cos x
Diff. bothsides w.r.t. x , we have
\(\frac{d y}{d x}\) = – a cosec² x + 0 + (- cosec x cot x) + xⁿ (- sin x) + cos x (n x^n-1)
[using product rule]
= – a cose² x – c cosec x cot x + x^n-1 (n cos x – x sin x)

(ii) Let y = \(\sqrt{\frac{\sec x+\tan x}{\sec x-\tan x}}\)
⇒ y = \(\sqrt{\frac{(\sec x+\tan x)(\sec x+\tan x)}{(\sec x-\tan x)(\sec x+\tan x)}}\)
⇒ y = sec x + tan x
[∵ sec² x – tan² x = 1]
Diff. bothsides w.r.t. x , we have
\(\frac{d y}{d x}\) = sec x tan x + sec² x
= sec x (tan x + sec x).

Question 8.
If y = \(\frac{2-3 \cos x}{\sin x}\), find \(\frac{d y}{d x}\) at x = \(\frac{\pi}{4}\).
Solution:
Given y = \(\frac{2-3 \cos x}{\sin x}\)
= 2 cosec x – 3 cot x
Diff. bothsides w.r.t. x , we have
\(\frac{d y}{d x}\) = – 2 cosec \(\frac{\pi}{4}\) cot \(\frac{\pi}{4}\) + 3 cosec² \(\frac{\pi}{4}\)
= – 2 × √2 × 1 + (3√2)²
= – 2√2 + 6
= 6 – 2√2.

Question 9.
If y = (sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\))², find \(\frac{d y}{d x}\) at x = \(\frac{\pi}{6}\).
Solution:
Given y = (sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\))²
⇒ y = sin² \(\frac{x}{2}\) + cos² \(\frac{x}{2}\) + 2 sin \(\frac{x}{2}\) cos \(\frac{x}{2}\)
⇒ y = 1 + sin x ;
Diff. bothsides w.r.t. x , we have
\(\frac{d y}{d x}\) = + cos x at x = \(\frac{\pi}{6}\),
\(\frac{d y}{d x}\) = cos \(\frac{\pi}{6}\)
= \(\frac{\sqrt{3}}{2}\).