NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.3

Mathematics is a science that deals with the Logic of Quantities, Shapes, and Arrangements. It is like a practice that goes everywhere around students. Therefore, Mathematics is a subject that students learn since their childhood.

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Chapter – 5 in Class 12 Mathematics is Continuity and Differentiability. The chapter includes the following topics – Continuity, Differentiability, Exponential and Logarithmic Functions, Logarithmic Differentiation, Derivatives of Functions in Parametric Forms, Second Order Derivatives and Mean Value Theorem. Hence, the chapter covers a wide range of topics with very important concepts.

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NCERT Solutions for Class 12th Maths Chapter 5 Continuity and Differentiability (Ex 5.3) Exercise 5.3

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Access NCERT Solutions for Mathematics Chapter 5 – Continuity and Differentiability

Important Points

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3 comprises the following topics –

Derivatives of Implicit Functions
Derivatives of Inverse Trigonometric Functions

Some important concepts of the Exercise 5.3 Class 12th are –

Sum, difference, product, and quotient of continuous functions are continuous.
Every differentiable function is continuous, but the converse is not true.
Rolle’s Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f(a) = f(b), then there exists some c in (a, b) such that f ′(c) = 0.
Mean Value Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that f'(c) = (f(b) – f(a))/(b-a).

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NCERT Solutions for Class 12 Maths Chapter 5 Other Exercises

Chapter 5 – Continuity and Differentiability Exercises
Exercise 5.1	10 Short Questions and 24 Long Questions
Exercise 5.3	9 short Questions and 6 long Questions
Exercise 5.4	5 Short Questions and 5 Long Questions
Exercise 5.5	4 Short Questions and 14 Long Questions
Exercise 5.6	1 Short Question and 1 Long Question
Exercise 5.7	10 Short Questions and 7 Long Questions
Exercise 5.8	Questions with Solutions

Q.1

Find dy/dx in the following:
2x + 3y = sinx

Ans

\begin{array}{l} Given : 2 x + 3 y = sinx \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (2 x + 3 y) = \frac{d}{dx} sinx \\ 2 \frac{d}{dx} x + 3 \frac{dy}{dx} = cosx \\ 2 + 3 \frac{dy}{dx} = cosx \\ \frac{dy}{dx} = \frac{cosx - 2}{3} \end{array}

Q.2

Find dy/dx in the following:
2x + 3y = siny

Ans

\begin{array}{l} Given : 2 x + 3 y = siny \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (2 x + 3 y) = \frac{d}{dx} siny \\ 2 \frac{d}{dx} x + 3 \frac{dy}{dx} = cosy \frac{dy}{dx} \\ 2 + 3 \frac{dy}{dx} = cosy \frac{dy}{dx} \\ 3 \frac{dy}{dx} - cosy \frac{dy}{dx} = - 2 \\ \frac{dy}{dx} (3 - cosy) = - 2 \\ \frac{dy}{dx} = \frac{- 2}{(3 - cosy)} \\ ∴ \frac{dy}{dx} = \frac{2}{(cosy - 3)} \end{array}

Q.3

Find \frac{dy}{dx} in the following : ax + {by}^{2} = cosy

Ans

\begin{array}{l} Given : ax + {by}^{2} = cosy \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (ax + {by}^{2}) = \frac{d}{dx} cosy \\ a \frac{d}{dx} x + 2 by \frac{dy}{dx} = - siny \frac{dy}{dx} \\ a + 2 by \frac{dy}{dx} = - siny \frac{dy}{dx} \\ 2 by \frac{dy}{dx} + siny \frac{dy}{dx} = - a \\ \frac{dy}{dx} (2 by + siny) = - a \\ \frac{dy}{dx} = \frac{- a}{(2 by + siny)} \\ ∴ \frac{dy}{dx} = \frac{- a}{(2 by + siny)} \end{array}

Q.4

Find \frac{dy}{dx} in the following : xy + y^{2} = tanx + y

Ans

\begin{array}{l} Given : xy + y^{2} = tanx + y \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (xy + y^{2}) = \frac{d}{dx} (tanx + y) \\ [x \frac{dy}{dx} + y \frac{d}{dx} x] + \frac{d}{dx} y^{2} = \sec^{2} x + \frac{dy}{dx} [By ​ Product Rule] \\ x \frac{dy}{dx} + y + 2 y \frac{dy}{dx} = \sec^{2} x + \frac{dy}{dx} \\ x \frac{dy}{dx} + 2 y \frac{dy}{dx} - \frac{dy}{dx} = \sec^{2} x - y \\ (x + 2 y - 1) \frac{dy}{dx} = \sec^{2} x - y \\ \frac{dy}{dx} = \frac{\sec^{2} x - y}{(x + 2 y - 1)} \end{array}

Q.5

Find \frac{dy}{dx} in the following x^{3} + x^{2} y + {xy}^{2} + y^{3} = 81

Ans

\begin{array}{l} Given : x^{3} + x^{2} y + {xy}^{2} + y^{3} = 81 \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (x^{3} + x^{2} y + {xy}^{2} + y^{3}) = \frac{d}{dx} 81 \\ \frac{d}{dx} x^{3} + \frac{d}{dx} x^{2} y + \frac{d}{dx} {xy}^{2} + \frac{d}{dx} y^{3} = 0 \\ 3 x^{2} + (x^{2} \frac{dy}{dx} + y \frac{d}{dx} x^{2}) + (x \frac{d}{dx} y^{2} + y^{2} \frac{d}{dx} x) + 3 y^{2} \frac{dy}{dx} = 0 \\ [By product rule] \\ 3 x^{2} + x^{2} \frac{dy}{dx} + y \times 2 x + (2 xy \frac{dy}{dx} + y^{2} \times 1) + 3 y^{2} \frac{dy}{dx} = 0 \\ 3 x^{2} + x^{2} \frac{dy}{dx} + 2 xy + 2 xy \frac{dy}{dx} + y^{2} + 3 y^{2} \frac{dy}{dx} = 0 \\ (x^{2} + 2 xy + 3 y^{2}) \frac{dy}{dx} = - 3 x^{2} - 2 xy - y^{2} \\ ∴ \frac{dy}{dx} = - \frac{(3 x^{2} + 2 xy + y^{2})}{(x^{2} + 2 xy + 3 y^{2})} \end{array}

Q.6

\begin{array}{l} Find \frac{dy}{dx} in the following \\ x^{2} + xy + y^{2} = 100 \end{array}

Ans

\begin{array}{l} Given : x^{2} + xy + y^{2} = 100 \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (x^{2} + xy + y^{2}) = \frac{d}{dx} 100 \\ \frac{d}{dx} x^{2} + \frac{d}{dx} xy + \frac{d}{dx} y^{2} = 0 \\ 2 x + (x \frac{dy}{dx} + y \frac{d}{dx} x) + 2 y \frac{dy}{dx} = 0 [By product rule] \\ 2 x + x \frac{dy}{dx} + y \times 1 + 2 y \frac{dy}{dx} = 0 \\ \frac{dy}{dx} (x + 2 y) = - 2 x - y \\ \frac{dy}{dx} = - \frac{2 x + y}{x + 2 y} \end{array}

Q.7

{sin}^{2} y + cosxy = K

Ans

\begin{array}{l} Given : \sin^{2} y + cosxy = K \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (\sin^{2} y + cosxy) = \frac{d}{dx} K \\ \frac{d}{dx} \sin^{2} y + \frac{d}{dx} cosxy = 0 \\ 2 siny \frac{d}{dx} siny - sinxy \frac{d}{dx} (xy) = 0 [Using chain rule] \\ 2 sinycosy \frac{dy}{dx} - sinxy (x \frac{dy}{dx} + y \frac{d}{dx} x) = 0 [By product rule] \\ \sin 2 y \frac{dy}{dx} - xsinxy \frac{dy}{dx} - ysinxy = 0 \\ \frac{dy}{dx} (\sin 2 y - xsinxy) = ysinxy \\ \frac{dy}{dx} = \frac{ysinxy}{(\sin 2 y - xsinxy)} \end{array}

Q.8

\begin{array}{l} Find \frac{dy}{dx} in the following \\ \sin^{2} x + \cos^{2} y = 1 \end{array}

Ans

\begin{array}{l} Given : \sin^{2} x + \cos^{2} y = 1 \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (\sin^{2} x + \cos^{2} y) = \frac{d}{dx} 1 \\ \Rightarrow \frac{d}{dx} \sin^{2} x + \frac{d}{dx} \cos^{2} y = 0 \\ \Rightarrow 2 sinx \frac{d}{dx} sinx + 2 cosy \frac{d}{dx} (cosy) = 0 [Using chain rule] \\ \Rightarrow 2 sinxcosx + 2 cosy (- siny) \frac{dy}{dx} = 0 \\ \Rightarrow \sin 2 x - \sin 2 y \frac{dy}{dx} = 0 \\ \Rightarrow - \sin 2 y \frac{dy}{dx} = - \sin 2 x \\ \Rightarrow \frac{dy}{dx} = \frac{\sin 2 x}{\sin 2 y} \end{array}

Q.9

Find \frac{dy}{dx} in the following y = \sin^{- 1} (\frac{2 x}{1 + x^{2}})

Ans

\begin{array}{l} The given relationship is \\ y = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \\ \Rightarrow siny = \frac{2 x}{1 + x^{2}} \\ Differentiating both sides w . r . t . x, we get \\ \Rightarrow \frac{d}{dx} siny = \frac{d}{dx} (\frac{2 x}{1 + x^{2}}) \\ \Rightarrow cosy \frac{dy}{dx} = \frac{(1 + x^{2}) \frac{d}{dx} 2 x - 2 x \frac{d}{dx} (1 + x^{2})}{{(1 + x^{2})}^{2}} \\ [By Quotient Rule] \\ \Rightarrow \sqrt{1 - \sin^{2} y} \frac{dy}{dx} = \frac{(1 + x^{2}) \times 2 - 2 x (0 + 2 x)}{{(1 + x^{2})}^{2}} \\ \Rightarrow \sqrt{1 - {(\frac{2 x}{1 + x^{2}})}^{2}} \frac{dy}{dx} = \frac{2 + 2 x^{2} - 4 x^{2}}{{(1 + x^{2})}^{2}} [Putting value of siny] \\ \Rightarrow \sqrt{\frac{{(1 + x^{2})}^{2} - 4 x^{2}}{{(1 + x^{2})}^{2}}} \frac{dy}{dx} = \frac{2 - 2 x^{2}}{{(1 + x^{2})}^{2}} \\ \Rightarrow \sqrt{\frac{{(1 - x^{2})}^{2}}{{(1 + x^{2})}^{2}}} \frac{dy}{dx} = \frac{2 - 2 x^{2}}{{(1 + x^{2})}^{2}} \\ \Rightarrow \frac{(1 - x^{2})}{(1 + x^{2})} \frac{dy}{dx} = \frac{2 - 2 x^{2}}{{(1 + x^{2})}^{2}} \\ \Rightarrow \frac{dy}{dx} = \frac{2 - 2 x^{2}}{(1 + x^{2}) (1 - x^{2})} \\ \Rightarrow \frac{dy}{dx} = \frac{2 (1 - x^{2})}{(1 + x^{2}) (1 - x^{2})} \\ ∴ \frac{dy}{dx} = \frac{2}{(1 + x^{2})} \end{array}

Q.10

Find \frac{dy}{dx} in the following : y = \tan^{- 1} (\frac{3 x - x^{3}}{1 + 3 x^{2}}), - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}

Ans

\begin{array}{l} We ​​ have y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) \\ Putting x = tanθ, ​ we get \\ y = \tan^{- 1} (\frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) \\ = \tan^{- 1} (\tan 3 θ) \\ Since, - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}, then \\ x = tanθ \\ \Rightarrow - \frac{1}{\sqrt{3}} < tanθ < \frac{1}{\sqrt{3}} \\ \Rightarrow - \frac{π}{6} < θ < \frac{π}{6} \\ \Rightarrow - \frac{π}{2} < 3 θ < \frac{π}{2} \\ ∴ y = 3 θ [∵ - \frac{π}{2} < 3 θ < \frac{π}{2}] \\ = 3 \tan^{- 1} x [∵ x = tanθ \Rightarrow θ = \tan^{- 1} x] \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = 3 \frac{d}{dx} \tan^{- 1} x \\ = 3 \times \frac{1}{1 + x^{2}} \\ = \frac{3}{1 + x^{2}} \\ ∴ \frac{d}{dx} \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) = \frac{3}{1 + x^{2}} \end{array}

Q.11

Find \frac{dy}{dx} in the following : y = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1

Ans

\begin{array}{l} We ​​ have y = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1 \\ Putting x = tanθ, ​ we get \\ y = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \\ = \cos^{- 1} (\cos 2 θ) \\ Since, 0 < x < 1, then \\ x = tanθ \\ \Rightarrow 0 < tanθ < 1 \\ \Rightarrow 0 < θ < \frac{π}{4} \\ \Rightarrow 0 < 2 θ < \frac{π}{2} \\ ∴ y = 2 θ [∵ 0 < 2 θ < \frac{π}{2}] \\ = 2 \tan^{- 1} x [∵ x = tanθ \Rightarrow θ = \tan^{- 1} x] \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = 2 \frac{d}{dx} \tan^{- 1} x \\ = 2 \times \frac{1}{1 + x^{2}} \\ = \frac{2}{1 + x^{2}} \\ ∴ \frac{d}{dx} \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) = \frac{2}{1 + x^{2}} \end{array}

Q.12

Find \frac{dy}{dx} in the following : y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1

Ans

\begin{array}{l} We ​​ have y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1 \\ Putting x = tanθ, ​ we get \\ y = \sin^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \\ = \sin^{- 1} (\cos 2 θ) \\ = \sin^{- 1} {\sin (\frac{π}{2} - 2 θ)} \\ Since, 0 < x < 1, then \\ x = tanθ \\ \Rightarrow 0 < tanθ < 1 \\ \Rightarrow 0 < θ < \frac{π}{4} \\ \Rightarrow 0 < 2 θ < \frac{π}{2} \\ \Rightarrow 0 < \frac{π}{2} - 2 θ < \frac{π}{2} \\ ∴ y = \frac{π}{2} - 2 θ [∵ 0 < \frac{π}{2} - 2 θ < \frac{π}{2}] \\ = \frac{π}{2} - 2 \tan^{- 1} x [∵ x = tanθ \Rightarrow θ = \tan^{- 1} x] \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = 0 - 2 \frac{d}{dx} \tan^{- 1} x \\ = - 2 \times \frac{1}{1 + x^{2}} \\ = - \frac{2}{1 + x^{2}} \\ Thus, \\ \frac{d}{dx} \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = - \frac{2}{1 + x^{2}} \end{array}

Q.13

Find \frac{dy}{dx} in the following : y = \cos^{- 1} (\frac{2 x}{1 + x^{2}}), - 1 < x < 1

Ans

\begin{array}{l} We ​​ have y = \cos^{- 1} (\frac{2 x}{1 + x^{2}}), - 1 < x < 1 \\ Putting x = tanθ, ​ we get \\ y = \cos^{- 1} (\frac{2 tanθ}{1 + \tan^{2} θ}) \\ = \cos^{- 1} (\sin 2 θ) \\ = \cos^{- 1} {\cos (\frac{π}{2} - 2 θ)} \\ Since, - 1 < x < 1, then \\ x = tanθ \\ \Rightarrow - 1 < tanθ < 1 \\ \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} \\ \Rightarrow - \frac{π}{2} < 2 θ < \frac{π}{2} \\ \Rightarrow \frac{π}{2} > - 2 θ > - \frac{π}{2} \\ \Rightarrow π > \frac{π}{2} - 2 θ > 0 \\ \Rightarrow 0 < \frac{π}{2} - 2 θ < π \\ ∴ y = \frac{π}{2} - 2 θ \\ = \frac{π}{2} - 2 \tan^{- 1} x [∵ x = tanθ \Rightarrow θ = \tan^{- 1} x] \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \frac{π}{2} - 2 \frac{d}{dx} \tan^{- 1} x \\ = 0 - 2 \times \frac{1}{1 + x^{2}} \\ ∴ \frac{d}{dx} \cos^{- 1} (\frac{2 x}{1 + x^{2}}) \\ = - \frac{2}{1 + x^{2}} . \end{array}

Q.14

Find \frac{dy}{dx} in the following y = \sin^{- 1} (2 x \sqrt{1 - x^{2}}), - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}

Ans

\begin{array}{l} We ​​ have y = \sin^{- 1} (2 x \sqrt{1 - x^{2}}), - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \\ Putting x = sinθ, ​ we get \\ y = \sin^{- 1} (2 sinθ \sqrt{1 - \sin^{2} θ}) \\ = \sin^{- 1} (2 sinθcosθ) \\ = \sin^{- 1} (\sin 2 θ) \\ Since, - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}, then \\ x = sinθ \\ \Rightarrow - \frac{1}{\sqrt{2}} < sinθ < \frac{1}{\sqrt{2}} \\ \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} \\ \Rightarrow - \frac{π}{2} < 2 θ < \frac{π}{2} \\ ∴ y = 2 θ \\ = 2 \sin^{- 1} x [x = sinθ \Rightarrow θ = \sin^{- 1} x] \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (2 \sin^{- 1} x) \\ = \frac{2}{\sqrt{1 - x^{2}}} \\ Thus, \\ \frac{d}{dx} {\sin^{- 1} (2 x \sqrt{1 - x^{2}})} = \frac{2}{\sqrt{1 - x^{2}}} \end{array}

Q.15

Find dy/dx in the following

{y = sec}^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}}

Ans

\begin{array}{l} We have \\ y = \sec^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}} \\ = \cos^{- 1} (2 x^{2} - 1) \\ Putting x = cosθ, ​ we get \\ y = \cos^{- 1} (2 \cos^{2} θ - 1) \\ = \cos^{- 1} (\cos 2 θ) \\ Since, 0 < x < \frac{1}{\sqrt{2}} \\ \Rightarrow 0 < cosθ < \frac{1}{\sqrt{2}} [∵ x = cosθ] \\ \Rightarrow \frac{3 π}{2} < θ < \frac{7 π}{4} \\ \Rightarrow 0 < θ < \frac{π}{4} \\ \Rightarrow 0 < 2 θ < \frac{π}{2} \\ ∴ y = 2 θ \\ = 2 \cos^{- 1} x [∵ x = cosθ \Rightarrow θ = \cos^{- 1} x] \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = 2 \frac{d}{dx} \cos^{- 1} x \\ = 2 (- \frac{1}{\sqrt{1 - x^{2}}}) \\ = - \frac{2}{\sqrt{1 - x^{2}}} \\ Thus, \\ \frac{d}{dx} \sec^{- 1} (\frac{1}{2 x^{2} - 1}) = - \frac{2}{\sqrt{1 - x^{2}}} . \end{array}

NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.3

NCERT Solutions for Class 12 Chapter 5 Exercise 5.3

NCERT Solutions for Class 12th Maths Chapter 5 Continuity and Differentiability (Ex 5.3) Exercise 5.3

Access NCERT Solutions for Mathematics Chapter 5 – Continuity and Differentiability

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NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.3

NCERT Solutions for Class 12 Chapter 5 Exercise 5.3

NCERT Solutions for Class 12th Maths Chapter 5 Continuity and Differentiability (Ex 5.3) Exercise 5.3

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