ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.7

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.7

Find \(\frac{d y}{d x}\) in the following (1 to 4) :

Question 1.
(i) x – y = π (NCERT)
(ii) x² + y² = 25
Solution:
(i) Given x – y = π ………..(1)
Diff. (1) w.r.t. x, we have
\(\frac{d}{d x}\) (x – y) = \(\frac{d}{d x}\) (π)
1 – \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = 1

(ii) Given, x² + y² = 25
Diff. both sides w.r.t. x, we have
Taking y as a function of x
\(\frac{d}{d x}\) (x²) + \(\frac{d}{d x}\) y² = \(\frac{d}{d x}\) (25)
⇒ 2x + 2y \(\frac{d y}{d x}\) = 0
⇒ x + y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{x}{y}\).

Question 2.
(i) xy = c²
(ii) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (ISC 2016)
Solution:
(i) Given, xy = c²
Diff. both sides w.r.t. x, we have
\(\frac{d}{d x}\) (xy) = \(\frac{d}{d x}\) (c²
⇒ x \(\frac{d y}{d x}\) + y . 1 = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\)

(ii) Given \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 ;
Diff. both sides w.r.t. x, we have
\(\frac{d}{d x}\left(\frac{x^2}{a^2}\right)+\frac{d}{d x}\left(\frac{y^2}{b^2}\right)\) = 0
⇒ \(\frac{1}{a^2} 2 x+\frac{1}{b^2} 2 y \frac{d y}{d x}\) = 0
⇒ \(\frac{2 y}{b^2} \frac{d y}{d x}=\frac{-2 x}{a^2}\)
⇒ \(\frac{d y}{d x}=\frac{-b^2 x}{a^2 y}\)

Question 3.
(i) √x + √y = √c
(ii) x³ + y³ = 3axy (ISC 2020)
Solution:
(i) Given, √x + √y = √c
Diff. both sides w.r.t. x, we have
\(\frac{d}{d x} \sqrt{x}+\frac{1}{d y} \sqrt{y}=\frac{d}{d x} \sqrt{c}\)
⇒ \(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}=-\frac{\sqrt{y}}{\sqrt{x}}=-\sqrt{\frac{y}{x}}\)

(ii) Given, x³ + y³ = 3axy ………….(1)
Diff. both sides of eqn. (1) w.r.t. x;
3x² + 3y² \(\frac{d y}{d x}\) = 3a [x \(\frac{d y}{d x}\) + y . 1]
⇒ (y² – ax) \(\frac{d y}{d x}\) = ay – x²
⇒ \(\frac{d y}{d x}=\frac{a y-x^2}{y^2-a x}\)

Question 3 (old).
(ii) x³ + x²y + xy² + y³ = 81. (NCERT)
Solution:
Given, x³ + x²y + xy² + y³ = 81 …………..(1)
Diff. (1) w.r.t. x, we have
3x² + x² + y (2x) + x (2y \(\frac{d y}{d x}\)) + y² . 1 + 3y² \(\frac{d y}{d x}\) = 0
⇒ (x² + 2xy + y²) \(\frac{d y}{d x}\) = – (3x² + 2xy + y²)
⇒ \(\frac{d y}{d x}=-\frac{\left(3 x^2+2 x y+y^2\right)}{x^2+2 x y+3 y^2}\).

Question 4.
(i) 2x + 3y = sin x (NCERT)
(ii) 2x + 3y = sin y (NCERT)
Solution:
Given, 2x + 3y = sin x ;
Diff. both sides w.r.t. x, we have
2 + 3 \(\frac{d y}{d x}\) = cos x
⇒ \(\frac{d y}{d x}\) = \(\frac{\cos x-2}{3}\)

(ii) Given, 2x + 3y = sin y ;
Diff. both sides w.r.t. x, we have
2 + 3 \(\frac{d y}{d x}\) = cos y \(\frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{2}{\cos y-3}\)

Question 5.
Find \(\frac{d y}{d x}\) when
(i) (x² + y²)² = xy
(ii) sin (x + y) = \(\frac{2}{3}\)
Solution:
(i) Given (x² + y²)² = xy
Differentiating both sides w.r.t. x ; we get
2 (x² + y²) \(\frac{d}{d x}\) (x² + y²) = \(\frac{d}{d x}\) (xy)
⇒ 2 (x² + y²) [\(\frac{d}{d x}\) x² + \(\frac{d}{d x}\) y²] = x \(\frac{d y}{d x}\) + y \(\frac{d}{d x}\) (x)
⇒ 2 (x² + y²) [2x + 2y \(\frac{d y}{d x}\)] = x \(\frac{d y}{d x}\) + y × 1
⇒ [4y (x² + y²) – x] \(\frac{d y}{d x}\) = y – 4x (x² + y²)
⇒ \(\frac{d y}{d x}=\frac{4 x\left(x^2+y^2\right)-y}{x-4\left(x^2+y^2\right) y}\)

(ii) Given, sin (x + y) = \(\frac{2}{3}\) …………..(1)
Diff. (1) w.r.t. x ; we have
cos (x + y) . \(\frac{d}{d x}\) (x + y) = 0
⇒ cos (x + y) [1 + \(\frac{d y}{d x}\)] = 0
⇒ 1 + \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – 1 [∵ cos (x + y) ≠ 0]

Question 6.
(i) x^2/3 + y^2/3 = 2, find \(\frac{d y}{d x}\) at (1, 1).
Solution:
Given, x^2/3 + y^2/3 = 2 ………..(1)
Diff. (1) w.r.t. x ; we have
\(\frac{2}{3}\) x^-1/3 + \(\frac{2}{3}\) y^-1/3 \(\frac{d y}{d x}\) = 0
⇒ y^-1/3 \(\frac{d y}{d x}\) = – x^-1/3
⇒ \(\frac{d y}{d x}=\frac{-y^{1 / 3}}{x^{1 / 3}}\)
∴ at (1, 1) ; \(\frac{d y}{d x}\) = – \(\frac{1}{1}\) = – 1

Question 7.
Use implicit differentiation to verify that \(\frac{d y}{d x}\) . \(\frac{d x}{d y}\) = 1, when
(i) y² = 4 ax
(ii) x⁴ + y⁴ = 3x²y²
Solution:
(i) Given, y² = 4 ax ………………(1)
Diff. eqn. (1) w.r.t. x ; we have
(taking x as a function of y)
\(\frac{d}{d y}\) y² = \(\frac{d}{d y}\) (4ax)
⇒ 2y = 4a \(\frac{d x}{d y}\) ……………(2)
⇒ \(\frac{d x}{d y}=\frac{y}{2 a}\) …………….(3)
∴ \(\frac{d y}{d x} \cdot \frac{d x}{d y}=\frac{2 a}{y} \times \frac{y}{2 a}\) = 1 [using eqn. (2) and (3)]

(ii) Given, x⁴ + y⁴ = 3x²y² ………(1)
Diff. eqn. (1) bothsides w.r.t. x ; we have
4x³ + 4y³ \(\frac{d y}{d x}\) = 3 [x² . 2y \(\frac{d y}{d x}\) + y² . 2x]
⇒ (4y³ – 6x²y) \(\frac{d y}{d x}\) = 6xy² – 4x³

Question 7 (old).
(ii) x³ + y³ = 3 axy.
Solution:
Given, x³ + y³ = 3 axy ………….(1)
Diff. eqn. (1) w.r.t. x ; we have
\(\frac{d}{d x}\) x³ + \(\frac{d}{d x}\) y³ = \(\frac{d}{d x}\) 3axy
⇒ 3x² + 3y² \(\frac{d y}{d x}\) = 3a [x \(\frac{d y}{d x}\) + y]
⇒ 3 (y² – ax) \(\frac{d y}{d x}\) = 3 (ay – x²)
⇒ \(\frac{d y}{d x}=\frac{a y-x^2}{y^2-a x}\) ………….(2)
Diff. eqn. (1) w.r.t. y ; we have
(by taking x as a function of y)

Question 8.
Find \(\frac{d y}{d x}\), when
(i) y + sin y = cos x (NCERT)
(ii) ax + by² = cos y (NCERT)
(iii) y sec x + tan x + x² y = 0
(iv) sin² x + cos² y = 1 (NCERT)
(v) xy + y² = tan x + y
(vi) y = tan (x + y) (NCERT Exampler)
(vii) sin (xy) + \(\frac{x}{y}\) = x² – y (NCERT Exampler)
(viii) sec (x + y) = xy. (NCERT Exampler)
Solution:
(i) Given, y + sin y = cos x …………(1)
Diff. eqn. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) + cos y \(\frac{d y}{d x}\) = – sin x
⇒ (1 + cos y) \(\frac{d y}{d x}\) = – sin x
⇒ \(\frac{d y}{d x}\) = – \(\frac{\sin x}{1+\cos y}\)

(ii) Given, ax + by² = cos y
Diff. both sides w.r.t. x ; we have
a + 2by \(\frac{d y}{d x}\) = – sin y \(\frac{d y}{d x}\)
⇒ (2xy + sin y )\(\frac{d y}{d x}\) = – a
⇒ \(\frac{d y}{d x}\) = \(-\frac{a}{2 b y+\sin y}\)

(iii) Given, y sec x + tan x + x² y = 0
Diff. both sides w.r.t. x ; we have
y sec x tan x + sec x \(\frac{d y}{d x}\) + sec² x + x² \(\frac{d y}{d x}\) + y . 2x = 0
⇒ (sec x + x²) \(\frac{d y}{d x}\) = – sec² x – y sec x tan x – 2xy
⇒ \(\frac{d y}{d x}=-\frac{2 x y+\sec ^2 x+y \sec x \tan x}{\left(\sec x+x^2\right)}\)

(iv) Given, sin² x + cos² y = 1
Diff. both sides w.r.t. x ; we have,
Taking y as a function of x, we get
2 sin x cos x + 2 cos y (- sin y) \(\frac{d y}{d x}\) = 0
⇒ sin 2x – sin 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{\sin 2 x}{\sin 2 y}\).

(v) Given, xy + y² = tan x + y
Diff. both sides w.r.t. x ; we have,
Taking y as a function of x, we have
x \(\frac{d y}{d x}\) + y . 1 + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{\sin 2 x}{\sin 2 y}\)

(vi) Given, y = tan (x + y)
Diff. both sides w.r.t. x ; we have,
\(\frac{d y}{d x}\) = sec² (x + y) [1 + \(\frac{d y}{d x}\)]
⇒ (1 – sec² (x + y)) \(\frac{d y}{d x}\) = sec² (x + y)
⇒ \(\frac{d y}{d x}=\frac{\sec ^2(x+y)}{1-\sec ^2(x+y)}\)
= \(\frac{\sec ^2(x+y)}{-\tan ^2(x+y)}\)
⇒ \(\frac{d y}{d x}=-\frac{1}{\sin ^2(x+y)}\)
= – cosec² (x + y)

(vii) Given, sin (xy) + \(\frac{y}{x}\) = x² – y² ………….(1)
Differentiating eqn. (1) both sides, we get
\(\frac{d}{d x}\) sin (xy) + \(\frac{d}{d x}\) (\(\frac{y}{x}\)) = \(\frac{d}{d x}\) x² – \(\frac{d}{d y}\) y²
⇒ cos (xy) \(\frac{d}{d x}\) (xy) + \(\frac{x \frac{d y}{d x}-y \times 1}{x^2}\)
= 2x – 2y \(\frac{d y}{d x}\)
⇒ x² cos (xy) [x \(\frac{d y}{d x}\) + y] + x \(\frac{d y}{d x}\) – y = x² [2x – 2y \(\frac{d y}{d x}\)]
⇒ (x³ cos(xy) + x + 2x² y) \(\frac{d y}{d x}\) = 2x³ + y – x²y cos xy
⇒ \(\frac{d y}{d x}=\frac{2 x^3+y-x^2 y \cos x y}{x\left[x^2 \cos x y+1+2 x y\right]}\)

(viii) Given, sec (x + y) = xy
Diff. both sides w.r.t. x,
Taking y as a function of x, we have
sec (x + y) tan (x + y) [1 + \(\frac{d y}{d x}\)] = x \(\frac{d y}{d x}\) + y
⇒ [sec (x + y) tan (x + y) – x] \(\frac{d y}{d x}\) = y – sec (x + y) tan (x + y)
⇒ \(\frac{d y}{d x}\) = \(\frac{y-\sec (x+y) \tan (x+y)}{\sec (x+y) \tan (x+y)-x}\)

Question 9.
If sec \(\left(\frac{x+y}{x-y}\right)\) = a, prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Given, sec \(\left(\frac{x+y}{x-y}\right)\) = a
⇒ \(\frac{x+y}{x-y}\) = sec^-1 a = K (say)
applying componendo and dividendo, we have
\(\frac{x+y+x-y}{x+y-x+y}=\frac{\mathrm{K}+1}{\mathrm{~K}-1}\)
⇒ \(\frac{2 x}{2 y}=\frac{\mathrm{K}+1}{\mathrm{~K}-1}\)
⇒ \(\frac{y}{x}=\frac{\mathrm{K}-1}{\mathrm{~K}+1}\)
Diff. both sides w.r.t. x, we get
\(\frac{x \frac{d y}{d x}-y \cdot 1}{x^2}\) = 0
⇒ x \(\frac{d y}{d x}\) = y
⇒ \(\frac{d y}{d x}\) = \(\frac{y}{x}\)

Question 10.
(i) If y = x sin (a + y), prove that \(\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin (a+y)-y \cos (a+y)}\).
(ii) If sin x = y sin (x + a), prove that \(\frac{d y}{d x}=\frac{\sin a}{\sin ^2(x+a)}\).
(iii) If sin y = x sin (a + y), prove that \(\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}\).
Solution:
(i) Given y = x sin (a + y)
Differentiating both sides w.r.t. y, we get
⇒ \(\frac{d x}{d y}=\frac{\sin (a+y) \frac{d y}{d y}-y \frac{d}{d y} \sin (a+y)}{\sin ^2(a+y)}\)
⇒ \(\frac{d x}{d y}=\frac{\sin (a+y) \cdot 1-y \cos (a+y) \frac{d}{d y}(a+y)}{\sin ^2(a+y)}\)
∴ \(\frac{d x}{d y}=\frac{\sin (a+y)-y \cos (a+y)}{\sin ^2(a+y)}\)
Thus \(\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}\)
= \(\frac{\sin ^2(a+y)}{\sin (a+y)-y \cos (a+y)}\)

(ii) Given sin x = y sin (x + a)
⇒ y = \(\frac{\sin x}{\sin (x+a)}\)
Differentiating both sides w.r.t. y, we get
\(\frac{d y}{d x}=\frac{\sin (x+a) \frac{d}{d x} \sin x-\sin x \frac{d}{d x} \sin (x+a)}{\sin ^2(x+a)}\)
= \(\frac{\sin (x+a) \cos x-\sin x \cos (x+a)}{\sin ^2(x+a)}\)
= \(\frac{\sin (x+a-x)}{\sin ^2(x+a)}\)
[∵ sin A cos B – cos A sin B = sin (A – B)]
⇒ \(\frac{d y}{d x}=\frac{\sin a}{\sin ^2(x+a)}\)

(iii) Given, sin y = x sin (a + y)
⇒ x = \(\frac{\sin y}{\sin (a+y)}\) ……………(1)
Diff. both sides of eqn. (1) w.r.t. x,
Taking y as a function of x, we have
\(\frac{d x}{d y}=\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin ^2(a+y)}\)
= \(\frac{\sin (a+y-y)}{\sin ^2(a+y)}\)
[∵ sin A cos B – cos A sin B = sin (A – B)]
∴ \(\frac{d y}{d x}=\frac{\sin a}{\sin ^2(a+y)}\)
∴ \(\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}\)
= \(\frac{\sin ^2(a+y)}{\sin a}\)

Question 11.
(i) If y = \(\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}\), prove that (2y – 1) \(\frac{d y}{d x}\) = 1.
(ii) If y = \(\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\ldots \infty}}}\), prove that (2y – 1) \(\frac{d y}{d x}\) = cos x.
(iii) If y = \(\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}}\), prove that (2y – 1) \(\frac{d y}{d x}\) = sec² x.
Solution:
(i) Given, y = \(\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}\) ;
on squaring, we have
y² = x + y
Differentiating both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 1 + \(\frac{d y}{d x}\)
⇒ (2y – 1) \(\frac{d y}{d x}\) = 1.

(ii) Given, y = \(\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\ldots \infty}}}\)
⇒ y = \(\sqrt{\sin x+y}\) ;
On squaring ; we have
y² = sin x + y ………..(1)
Differentiating eqn. (1) both sides ;
Taking y as a function of x, we have
2y \(\frac{d y}{d x}\) = cos x + \(\frac{d y}{d x}\)
⇒ (2y – 1) \(\frac{d y}{d x}\) = cos x

(iii) The given eqn. can be written as
y = \(\sqrt{\tan x+y}\) ;
On squaring we get
y² = tan x + y
Differentiating both sides w.r.t. y, we get ;
⇒ 2y \(\frac{d y}{d x}\) = sec² x + \(\frac{d y}{d x}\)
⇒ (2y – 1) \(\frac{d y}{d x}\) = sec² x