Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

Question 1.
2x + 3y = sin x
Solution:
2x + 3y = sin x.
Differentiating w.r.t. x,
2x + 3(frac { dy }{ dx }) = cos x. ∴ (frac { dy }{ dx }) = (frac { 1 }{ 3 })(cosx-2)
2x + 3y = sin y.

Question 2.
2x + 3 y = sin y
Solution:
Differentiating w.r.t. x,
2 + 3(frac { dy }{ dx }) = cos y (frac { dy }{ dx }) or (cos y – 3) (frac { dy }{ dx }) = 2
∴ (frac { dy }{ dx }) = (frac { 2 }{ cos y – 3 }).
ax + by² = cos y.

Question 3.
ax + by² = cos y
Solution:
Differentiating w.r.t. x,
a + 2by(frac { dy }{ dx }) = – sin y (frac { dy }{ dx })
or (2b + sin y) (frac { dy }{ dx }) = – a
∴ (frac { dy }{ dx }) = – (frac { a }{ 2b + sin y }).
xy + y² = tan x + y.

Question 4.
xy + y² = tan x + y
Solution:
Differentiating w.r.t. x,
(1.y + x (frac { dy }{ dx })) = sec² x + (frac { dy }{ dx })
or (x + 2y – 1) (frac { dy }{ dx }) = sec² x – y
∴ (frac { dy }{ dx }) = (frac{sec ^{2} x-y}{x+2 y-1}).

Question 5.
x² + xy + y² = 100
Solution:
x² + xy + y² = 100
Differentiating w.r.t. x,
2x + (1.y + x. (frac { dy }{ dx })) + 2y (frac { dy }{ dx }) = 0
or (x + 2y) (frac { dy }{ dx }) = – 2x – y
∴ (frac { dy }{ dx }) = – (frac{ 2x+y }{ x+2y }).

Question 6.
x³ + x²y + xy² + y³ = 81
Solution:
x³ + x²y + xy² + y³ = 81
Differentiating w.r.t. x,

Question 7.
sin² y + cos xy = π
Solution:
sin² y + cos xy = π
Differentiating w.r.t. x,

Question 8.
sin² x + cos² y = 1
Solution:
sin² x + cos² y = 1
Differentiating w.r.t. x,
2 sin x (frac { d }{ dx }) (sin x) + 2 cosy (frac { d }{ dx }) (cos y) = 0
or 2 sin x cos x + 2 cos y(- sin y) (frac { dy }{ dx }) = 0
or 2sin x cos x – 2cos y sin y (frac { dy }{ dx }) = 0
∴ (frac { dy }{ dx }) = (frac { 2 sin x cos x}{ 2sin y cos y }) = (frac { sin 2x }{ sin 2y }).

Question 9.
y = sin^-1(left(frac{2 x}{1+x^{2}}right))
Solution:
y = sin^-1(left(frac{2 x}{1+x^{2}}right)). put x = tanθ
So, y = sin^-1(left(frac{2 tan theta}{1+tan ^{2} theta}right)) = sin^-1(sin 2θ)
= 2θ = 2 tan^-1x.

Question 10.
y = tan^-1(frac { 2x }{ 2 })(frac{-1}{sqrt{3}}) < x < (frac{1}{sqrt{3}})
Solution:
y = tan^-1(left(frac{3 x-x^{3}}{1-3 x^{2}}right)). put x = tanθ
So, y = tan^-1(left(frac{3 tan theta-tan ^{3} theta}{1-3 tan ^{2} theta}right)) = tan^-1(tan 3θ)
= 3θ = 3 tan^-1x.
∴ (frac { dy }{ dx }) = (frac{3}{1+x^{2}})

Question 11.
y = cos^-1(left(frac{3 x-x^{3}}{1-3 x^{2}}right)), 0 < x < 1
Solution:
y = cos^-1(left(frac{1-x^{2}}{1+x^{2}}right)). put x = tanθ
So, y = cos^-1(left(frac{1-tan ^{2} theta}{1+tan ^{2} theta}right)) = cos^-1(cos 2θ)
= 2θ = 2 tan^-1x.
∴ (frac { dy }{ dx }) = (frac{2}{1+x^{2}})

Question 12.
y = sin^-1(left(frac{1-x^{2}}{1+x^{2}}right)), 0 < x < 1
Solution:

Question 13.
y = cos^-1(left(frac{2 x}{1+x^{2}}right)), – 1 < x < 1
Solution:

Question 14.
y = sin^-1(left(2 x sqrt{1-x^{2}}right)), – (frac{-1}{sqrt{2}}) < x < (frac{1}{sqrt{2}})
Solution:
y = sin^-1(left(2 x sqrt{1-x^{2}}right))Put x = sin θ.
y = sin^-1(2sinθ(left.sqrt{1-sin ^{2} theta}right)))
= sin^-1(2 sin θ cos θ)
= sin^-1(sin 2θ) = 2θ = 2 ^-1x.
∴ (frac { dy }{ dx }) = (frac{2}{sqrt{1-x^{2}}})

Question 15.
y = sec^-1(left(frac{1}{2 x^{2}-1}right)), 0 < x < (frac{1}{sqrt{2}})
Solution:
y = sec^-1(left(frac{1}{2 x^{2}-1}right))Put x = cos θ.
y = sec^-1(left(frac{1}{2 cos ^{2} theta-1}right)) = sec^-1(frac{1}{cosθ})
= sec^-1(sec 2θ) = θ = 2 cos^-1x.
= sec^-1(sec 2θ) = 2θ = 2cos^-1x.
∴ (frac { dy }{ dx }) = (frac{- 2}{sqrt{1-x^{2}}})