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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 = nxn-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) $$\frac{d}{d x}$$ x1/2
= $$\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{2 \cos \sqrt{x} \sin \sqrt{x}}{2 \sqrt{x}}$$
= $$\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin (2 \sqrt{x})}{2 \sqrt{x}}$$
[∵ 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) sinm x cosn x (NCERT Exampler)
(ii) sinn (ax2 + bx + c) (NCERT Exampler)
Solution:
(i) Let y = sinm x cosn x ;
diff. bothsides w.r.t. x, we get
$$\frac{d y}{d x}$$ = sinm x $$\frac{d}{d x}$$ (cos x)n + cosn x $$\frac{d}{d x}$$ (sin x)m
= sinm x . n (cos x)n-1 (- sin x) + cosn x . m (sin x)m-1 . cos x
= – n sinm+1 x cosn-1 x + m (sin x)m-1 cosn+1 x
= sinn-1 x cosn-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 sinn-1 (ax2 + bx + c) cos (ax2 + bx + c) $$\frac{d}{d x}$$ (ax2 + bx + c)
= n sinn-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 + 4x3/2
(ii) 3 cot x + 6 (1 – 2x)5/3
Solution:
(i) Let y = 5 tan x – 3 sin x + 4x3/2
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 xn-1)
[using product rule]
= – a cose2 x – c cosec x cot x + xn-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 $$\frac{\pi}{6}$$
= $$\frac{\sqrt{3}}{2}$$.