# NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.3

## NCERT Solutions for Class 12 Chapter 5 Exercise 5.3

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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

Students can download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3 from Extramarks’ website. These solutions are easily accessible and are in PDF format.

### 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

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

Q.2

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

Ans

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

Q.3

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

Ans

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

Q.4

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

Ans

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

Q.5

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

Ans

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

Q.6

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

Ans

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

Q.7

${\text{sin}}^{2}\mathrm{y}+\mathrm{cosxy}=\mathrm{K}$

Ans

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

Q.8

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

Ans

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

Q.9

$\text{Find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{in}\mathrm{the}\mathrm{following}\phantom{\rule{0ex}{0ex}}\mathrm{y}={\mathrm{sin}}^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}\right)$

Ans

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

Q.10

$\text{Find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{in}\mathrm{the}\mathrm{following}:\phantom{\rule{0ex}{0ex}}\mathrm{y}={\mathrm{tan}}^{–1}\left(\frac{3\mathrm{x}-{\mathrm{x}}^{3}}{1+3{\mathrm{x}}^{2}}\right),-\frac{1}{\sqrt{3}}<\mathrm{x}<\frac{1}{\sqrt{3}}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{y}={\mathrm{tan}}^{-1}\left(\frac{3\mathrm{x}-{\mathrm{x}}^{3}}{1-3{\mathrm{x}}^{2}}\right)\\ \mathrm{Putting}\mathrm{x}=\mathrm{tan\theta },\mathrm{}\mathrm{we}\mathrm{get}\\ \mathrm{y}={\mathrm{tan}}^{-1}\left(\frac{3\mathrm{tan\theta }-{\mathrm{tan}}^{3}\mathrm{\theta }}{1-3{\mathrm{tan}}^{2}\mathrm{\theta }}\right)\\ ={\mathrm{tan}}^{-1}\left(\mathrm{tan}3\mathrm{\theta }\right)\\ \mathrm{Since},\mathrm{ }-\frac{1}{\sqrt{3}}<\mathrm{x}<\frac{1}{\sqrt{3}},\mathrm{then}\\ \mathrm{ }\mathrm{x}=\mathrm{tan\theta }\\ ⇒ \mathrm{ }-\frac{1}{\sqrt{3}}<\mathrm{tan\theta }<\frac{1}{\sqrt{3}}\\ ⇒-\frac{\mathrm{\pi }}{6}<\mathrm{\theta }<\frac{\mathrm{\pi }}{6}\\ ⇒-\frac{\mathrm{\pi }}{2}<3\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\\ \therefore \mathrm{ }\mathrm{y}=3\mathrm{\theta }\left[\because -\frac{\mathrm{\pi }}{2}<3\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\right]\\ =3{\mathrm{tan}}^{-1}\mathrm{x}\left[\because \mathrm{x}=\mathrm{tan\theta }⇒\mathrm{\theta }={\mathrm{tan}}^{-1}\mathrm{x}\right]\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=3\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\\ =3×\frac{1}{1+{\mathrm{x}}^{2}}\\ =\frac{3}{1+{\mathrm{x}}^{2}}\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\left(\frac{3\mathrm{x}-{\mathrm{x}}^{3}}{1-3{\mathrm{x}}^{2}}\right)=\frac{3}{1+{\mathrm{x}}^{2}}\\ \end{array}$

Q.11

$\text{Find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{in}\mathrm{the}\mathrm{following}:\mathrm{y}={\mathrm{cos}}^{–1}\left(\frac{1-{\mathrm{x}}^{2}}{1+{\mathrm{x}}^{2}}\right), \mathrm{ }\mathbf{0}<\mathbf{x}<\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{y}={\mathrm{cos}}^{–1}\left(\frac{\mathrm{1}-{\mathrm{x}}^{\mathrm{2}}}{1+{\mathrm{x}}^{\mathrm{2}}}\right), \mathrm{ }0<\mathrm{ }\mathrm{x}< 1\\ \mathrm{Putting}\mathrm{x}=\mathrm{tan\theta },\mathrm{}\mathrm{we}\mathrm{get}\\ \mathrm{y}={\mathrm{cos}}^{–1}\left(\frac{1-{\mathrm{tan}}^{2}\mathrm{\theta }}{1+{\mathrm{tan}}^{2}\mathrm{\theta }}\right)\\ ={\mathrm{cos}}^{–1}\left(\mathrm{cos}2\mathrm{\theta }\right)\\ \mathrm{Since},\mathrm{ }0< \mathrm{x} < \mathrm{1},\mathrm{then}\\ \mathrm{ }\mathrm{x}=\mathrm{tan\theta }\\ ⇒ 0<\mathrm{tan\theta }<1\\ ⇒ \mathrm{ }0<\mathrm{\theta }<\frac{\mathrm{\pi }}{4}\\ ⇒ \mathrm{ }0<2\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\\ \therefore \mathrm{y}=2\mathrm{\theta }\left[\because 0<2\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\right]\\ =2{\mathrm{tan}}^{-1}\mathrm{x}\left[\because \mathrm{x}=\mathrm{tan\theta }⇒\mathrm{\theta }={\mathrm{tan}}^{-1}\mathrm{x}\right]\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=2\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\\ =2×\frac{1}{1+{\mathrm{x}}^{2}}\\ =\frac{2}{1+{\mathrm{x}}^{2}}\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{cos}}^{–1}\left(\frac{1-{\mathrm{tan}}^{2}\mathrm{\theta }}{1+{\mathrm{tan}}^{2}\mathrm{\theta }}\right)=\frac{2}{1+{\mathrm{x}}^{2}}\end{array}$

Q.12

$\text{Find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{in}\mathrm{the}\mathrm{following}:\mathrm{y}={\mathrm{sin}}^{–1}\left(\frac{1-{\mathrm{x}}^{2}}{1+{\mathrm{x}}^{2}}\right), \mathrm{ }\mathbf{0}<\mathbf{x}<\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{y}={\mathrm{sin}}^{–1}\left(\frac{\mathrm{1}-{\mathrm{x}}^{\mathrm{2}}}{1+{\mathrm{x}}^{\mathrm{2}}}\right), \mathrm{ }0<\mathrm{ }\mathrm{x}< 1\\ \mathrm{Putting}\mathrm{x}=\mathrm{tan\theta },\mathrm{}\mathrm{we}\mathrm{get}\\ \mathrm{y}={\mathrm{sin}}^{–1}\left(\frac{1-{\mathrm{tan}}^{2}\mathrm{\theta }}{1+{\mathrm{tan}}^{2}\mathrm{\theta }}\right)\\ ={\mathrm{sin}}^{–1}\left(\mathrm{cos}2\mathrm{\theta }\right)\\ ={\mathrm{sin}}^{–1}\left\{\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }\right)\right\}\\ \mathrm{Since},\mathrm{ }0< \mathrm{x} < \mathrm{1},\mathrm{then}\\ \mathrm{x}=\mathrm{tan\theta }\\ ⇒ 0<\mathrm{tan\theta }<1\\ ⇒ \mathrm{ }0<\mathrm{\theta }<\frac{\mathrm{\pi }}{4}\\ ⇒ \mathrm{ }0<2\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\\ ⇒ 0<\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\\ \therefore \mathrm{y}=\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }\left[\because 0<\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\right]\\ =\frac{\mathrm{\pi }}{2}-2{\mathrm{tan}}^{-1}\mathrm{x}\left[\because \mathrm{x}=\mathrm{tan\theta }⇒\mathrm{\theta }={\mathrm{tan}}^{-1}\mathrm{x}\right]\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=0-2\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\\ =-2×\frac{1}{1+{\mathrm{x}}^{2}}\\ =-\frac{2}{1+{\mathrm{x}}^{2}}\\ \mathrm{Thus},\\ \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{–1}\left(\frac{\mathrm{1}-{\mathrm{x}}^{\mathrm{2}}}{1+{\mathrm{x}}^{\mathrm{2}}}\right)=-\frac{2}{1+{\mathrm{x}}^{2}}\end{array}$

Q.13

$\text{Find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{in}\mathrm{the}\mathrm{following}:\phantom{\rule{0ex}{0ex}}\mathrm{y}={\mathrm{cos}}^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}\right), \mathrm{ }-\mathbf{1}<\mathbf{x}<\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{y}={\mathrm{cos}}^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{\mathrm{2}}}\right), \mathrm{ }-\mathrm{1}<\mathrm{x}<\mathrm{1}\\ \mathrm{Putting}\mathrm{x}=\mathrm{tan\theta },\mathrm{}\mathrm{we}\mathrm{get}\\ \mathrm{y}={\mathrm{cos}}^{–1}\left(\frac{2\mathrm{tan\theta }}{1+{\mathrm{tan}}^{2}\mathrm{\theta }}\right)\\ ={\mathrm{cos}}^{–1}\left(\mathrm{sin}2\mathrm{\theta }\right)\\ ={\mathrm{cos}}^{–1}\left\{\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }\right)\right\}\\ \mathrm{Since},\mathrm{ }-\mathrm{1}<\mathrm{x}<\mathrm{1},\mathrm{then}\\ \mathrm{x}=\mathrm{tan\theta }\\ ⇒ -\mathrm{1}<\mathrm{tan\theta }<1\\ ⇒ \mathrm{ }-\frac{\mathrm{\pi }}{4}<\mathrm{\theta }<\frac{\mathrm{\pi }}{4}\\ ⇒ \mathrm{ }-\frac{\mathrm{\pi }}{2}<2\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\\ ⇒ \frac{\mathrm{\pi }}{2}>-2\mathrm{\theta }>-\frac{\mathrm{\pi }}{2}\\ ⇒ \mathrm{\pi }>\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }>0\\ ⇒ 0<\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }<\mathrm{\pi }\\ \therefore \mathrm{y}=\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }\\ =\frac{\mathrm{\pi }}{2}-2{\mathrm{tan}}^{-1}\mathrm{x}\left[\because \mathrm{x}=\mathrm{tan\theta }⇒\mathrm{\theta }={\mathrm{tan}}^{-1}\mathrm{x}\right]\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\frac{\mathrm{\pi }}{2}-2\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\\ =0-2×\frac{1}{1+{\mathrm{x}}^{2}}\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{cos}}^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{\mathrm{2}}}\right)\\ =-\frac{2}{1+{\mathrm{x}}^{2}}.\end{array}$

Q.14

$\text{Find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{in}\mathrm{the}\mathrm{following}\phantom{\rule{0ex}{0ex}}\mathrm{y}={\mathrm{sin}}^{–1}\left(\mathbf{2}\mathbf{x}\sqrt{1-{\mathbf{x}}^{2}}\right), \mathrm{ }-\frac{1}{\sqrt{2}}<\mathbf{x}<\frac{1}{\sqrt{2}}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{y}={\mathrm{sin}}^{–1}\left(2\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}\right), \mathrm{ }-\frac{1}{\sqrt{2}}< \mathrm{x}<\mathrm{ }\frac{\mathrm{1}}{\sqrt{2}}\\ \mathrm{Putting}\mathrm{x}=\mathrm{sin\theta },\mathrm{}\mathrm{we}\mathrm{get}\\ \mathrm{y}={\mathrm{sin}}^{–1}\left(2\mathrm{sin\theta }\sqrt{1-{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta }}\right)\\ ={\mathrm{sin}}^{–1}\left(2\mathrm{sin\theta cos\theta }\right)\\ ={\mathrm{sin}}^{–1}\left(\mathrm{sin}2\mathrm{\theta }\right)\\ \mathrm{Since},\mathrm{ }-\frac{1}{\sqrt{2}}< \mathrm{x}<\mathrm{ }\frac{\mathrm{1}}{\sqrt{2}},\mathrm{then}\\ \mathrm{x}=\mathrm{sin\theta }\\ ⇒ \mathrm{ }-\frac{1}{\sqrt{2}}<\mathrm{sin\theta }<\frac{\mathrm{1}}{\sqrt{2}}\\ ⇒ \mathrm{ }-\frac{\mathrm{\pi }}{4}<\mathrm{\theta }<\frac{\mathrm{\pi }}{4}\\ ⇒ \mathrm{ }-\frac{\mathrm{\pi }}{2}<2\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\\ \therefore \mathrm{y}=2\mathrm{\theta }\\ =2{\mathrm{sin}}^{-1}\mathrm{x}\left[\mathrm{x}=\mathrm{sin\theta }⇒\mathrm{\theta }={\mathrm{sin}}^{-1}\mathrm{x}\right]\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\mathrm{sin}}^{-1}\mathrm{x}\right)\\ =\frac{2}{\sqrt{1-{\mathrm{x}}^{2}}}\\ \mathrm{Thus},\\ \frac{\mathrm{d}}{\mathrm{dx}}\left\{{\mathrm{sin}}^{-1}\left(2\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}\right)\right\}=\frac{2}{\sqrt{1-{\mathrm{x}}^{2}}}\end{array}$

Q.15

Find dy/dx in the following

${\text{y = sec}}^{–1}\left(\frac{1}{\mathbf{2}{\mathrm{x}}^{2}-\mathbf{1}}\right), \mathrm{ }\mathbf{0}<\mathbf{x}<\frac{1}{\sqrt{2}}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \mathrm{y}={\mathrm{sec}}^{–1}\left(\frac{\mathrm{1}}{2{\mathrm{x}}^{\mathrm{2}}–1}\right), 0<\mathrm{x}<\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\\ \mathrm{ }={\mathrm{cos}}^{-1}\left(2{\mathrm{x}}^{2}-1\right)\\ \mathrm{Putting}\mathrm{x}=\mathrm{cos\theta }, \mathrm{we}\mathrm{get}\\ \mathrm{y}={\mathrm{cos}}^{-1}\left(2{\mathrm{cos}}^{2}\mathrm{\theta }-1\right)\\ ={\mathrm{cos}}^{-1}\left(\mathrm{cos}2\mathrm{\theta }\right)\\ \mathrm{Since},\mathrm{ }0<\mathrm{ }\mathrm{x} <\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\\ ⇒ 0<\mathrm{ }\mathrm{cos\theta } \mathrm{<}\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left[\because \mathrm{x}=\mathrm{cos\theta }\right]\\ ⇒ \frac{3\mathrm{\pi }}{2}<\mathrm{\theta }<\frac{7\mathrm{\pi }}{4}\\ ⇒ 0<\mathrm{\theta }<\frac{\mathrm{\pi }}{4}\\ ⇒ 0<2\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\\ \therefore \mathrm{y}=2\mathrm{\theta }\\ \mathrm{ }=2{\mathrm{cos}}^{-1}\mathrm{x}\left[\because \mathrm{x}=\mathrm{cos\theta }⇒\mathrm{\theta }={\mathrm{cos}}^{-1}\mathrm{x}\right]\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=2\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{cos}}^{-1}\mathrm{x}\\ =2\left(-\frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\right)\\ =-\frac{2}{\sqrt{1-{\mathrm{x}}^{2}}}\\ \mathrm{Thus},\\ \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sec}}^{–1}\left(\frac{\mathrm{1}}{2{\mathrm{x}}^{\mathrm{2}}–1}\right)=-\frac{2}{\sqrt{1-{\mathrm{x}}^{2}}}.\end{array}$