GSEB Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

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.
∴ $\frac { dy }{ dx }$ = $\frac{- 2}{\sqrt{1-x^{2}}}$