NCERT Solutions for Chapter 5 Continuity and Differentiability Class 12 Maths Exercise 5.3

Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.3

Continuity and Differentiability Exercise 5.3 Solutions

1. Find dy/dx . 
2x + 3y = sin x 

Solution

The given relationship is 2x + 3y = sin x
Differentiating this relationship with respect to x, we obtain 

2. Find dy/dx 
2x + 3y = sin y 

Solution

The given relationship is 2x + 3y = sin y
Differentiating this relationship with respect to x, we obtain 

3. Find dy/dx
ax + by² = cos y

Solution

The given relationship is ax + by² = cos y 
Differentiating this relationship with respect to x, we obtain 

4. Find dy/dx 
xy + y² = tan x + y 

Solution

The given relationship is xy +  y² = tan x + y 
Differentiating this relationship with respect to x, we obtain 

5. Find dy/dx
x² + xy + y² = 100 

Solution

The given relationship is x² + xy + y² = 100 
Differentiating this relationship with respect to x, we obtain 

6. Find = dy/x 
x³ + x² y + xy² + y³ = 81

Solution

The given relationship is x³ + x² y + xy² + y³ = 81 
Differentiating this relationship with respect to x, we obtain 

7. Find dy/dx 
sin² y + cos xy = π

Solution

The given relationship is sin² y + cos xy = π
Differentiating this relationship with respect to x, we obtain 

From (1), (2) and (3), we obtain 
2 sin y cos y dy/dx - y sin xy - (x sin xy) dy/dx = 0
⇒ (2 sin y cos y - x sin xy) dy/dx = y sin xy 
⇒ (sin 2y - x sin xy) dy/dx = y sin xy 
∴ dy/dx = y sin xy/(sin 2y - x sin xy)

8. Find dy/dx
sin² x + cos² y = 1

Solution

The given relationship is sin² x + cos² y = 1 
Differentiating this relationship with respect to x, we obtain 

⇒ 2 sin x cos x + 2cos y(-sin y). dy/dx = 0 
⇒ sin 2x - sin 2y (dy/dx) = 0 
∴ dy/dx = sin 2x/sin 2y

9. Find dy/dx 
y = sin^-1 (2x/(1 + x² )

Solution

The given relationship is y = sin^-1 (2x/(1 + x² )
y = sin^-1 (2x/(1 + x² ) 
⇒ sin y = 2x/(1 + x² )
Differentiating this relationship with respect to x, we obtain 

The function, 2x/ (1 + x² ) ,  is of the form of u/v. 
Therefore, by quotient rule, we obtain 

10. Find dy/dx 
y = tan^-1 (3x - x³ )/(1 - 3x² ), -1/√3 < x < 1/√3 

Solution

The given relationship is y = tan^-1 (3x - x³ )/(1 - 3x² ),  

Comparing equations (1) and (2), we obtain 
x = tan y/3 
Differentiating this relationship with respect to x, we obtain 

11. Find dy/dx
y = cos^-1 (1 - x² )/(1 + x² ), 0 < x < 1

Solution

The given relationship is, 

On comparing L.H.S. and R.H.S of the above relationship, we obtain 
tan y/2 = x 
Differentiating this relationship with respect to x, we obtain 

12. Find dy/dx 

y = sin^-1 (1 - x² )/(1 + x² ), 0 < x < 1

Solution

The given relationship is y = sin^-1 (1 - x² )/(1 + x² )

y = sin^-1 (1 - x² )/(1 + x² ) 
⇒ sin y = (1 - x² )/(1 + x² ) 
Differentiating this relationship with respect to x, we obtain 
d/dx (sin y) = d/dx [(1 - x² )/(1 + x² ) ]  ...(1) 
Using chain rule, we obtain 
d/dx (sin y) = cos y . dy/dx 

13. Find dx/dy 

y = cos^-1 (2x)/(1 + x² ), 0 < x < 1

Solution

The given relationship is y = cos^-1 (2x)/(1 + x² ) 
y = cos^-1 (2x)/(1 + x² ) 
⇒ cos y = 2x/(1 + x² )
Differentiating this relationship with respect to x, we obtain 

14. Find dy/dx 
y = sin^-1 [2x√(1 - x² )], -1/√2 < x < 1/√2

Solution 

The given relationship is y = sin^-1 [2x√(1 - x² )] 
y = sin^-1 [2x√(1 - x² )]  
⇒ sin y = 2x√(1 - x² )
Differentiating this relationship with respect to x, we obtain 

15. Find dy/dx 
y =  sec^-1 [ 1/(2x² - 1), 0 < x < 1√2 

Solution

The given relationship is y = sec^-1 [ 1/(2x² - 1)
y = sec^-1 [ 1/(2x² - 1)
⇒ sec y = 1/(2x² - 1)
⇒ cos y = 2x² - 1 
⇒ 2x² = 1 + cos y 
⇒ 2x² = 2 cos² (y/2)
⇒ x = cos y/2 
Differentiating this relationship with respect to x, we obtain