# NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.3

Go back to  'Continuity and Differentiability'

## Chapter 5 Ex.5.3 Question 1

Find $$\frac{{dy}}{{dx}}\;\;:\;\;2x + 3y = \sin x$$

### Solution

Given, $$2x + 3y = \sin x$$

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dy}}\left( {2x + 3y} \right) &= \frac{d}{{dx}}\left( {\sin x} \right)\\ \Rightarrow\; \frac{d}{{dx}}\left( {2x} \right) + \frac{d}{{dx}}\left( {3y} \right) &= \cos x\\ \Rightarrow\; 2 + 3\frac{{dy}}{{dx}} &= \cos x\\ \Rightarrow\; 3\frac{{dy}}{{dx}} &= \cos x - 2\\ \therefore\; \frac{{dx}}{{dy}} &= \frac{{\cos x - 2}}{3}\end{align}

## Chapter 5 Ex.5.3 Question 2

Find $$\frac{{dy}}{{dx}}\;\;:\;\; 2x + 3y = \sin y$$

### Solution

Given, $$2x + 3y = \sin y$$

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( {2x} \right) + \frac{d}{{dx}}\left( {3y} \right) &= \frac{d}{{dx}}\left( {\sin y} \right)\\ \Rightarrow\; 2 + 3\frac{{dy}}{{dx}} &= \cos y\frac{{dy}}{{dx}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{By using chain rule}}} \right]\\ \Rightarrow\; 2 &= \left( {\cos y - 3} \right)\frac{{dy}}{{dx}}\\ \therefore\; \frac{{dy}}{{dx}} &= \frac{2}{{\cos y - 3}}\end{align}

## Chapter 5 Ex.5.3 Question 3

Find $$\frac{{dy}}{{dx}}\;\;: \;\;ax + b{y^2} = \cos y$$

### Solution

Given, $$ax + b{y^2} = \cos y$$

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( {ax} \right) + \frac{d}{{dx}}\left( {b{y^2}} \right) &= \frac{d}{{dx}}\left( {\cos y} \right)\\ \Rightarrow\; a + b\frac{d}{{dx}}\left( {{y^2}} \right) &= \frac{d}{{dx}}\left( {\cos y} \right) \qquad \qquad \quad\qquad \ldots \left( 1 \right) \\[5pt]\qquad\qquad\qquad\qquad \frac{d}{{dx}}\left( {{y^2}} \right) = 2y\frac{{dy}}{{dx}}\;\; and \;\;{\text{ }}&\frac{d}{{dx}}\left( {\cos y} \right) = - \sin y\frac{{dy}}{{dx}} \qquad\quad\ldots \left( 2 \right)\end{align}

From ($$1$$) and ($$2$$), we obtain

\begin{align} a + b \times 2y\frac{{dy}}{{dx}} &= - \sin y\frac{{dy}}{{dx}}\\ \Rightarrow\; \left( {2by + \sin y} \right)\frac{{dy}}{{dx}} &= - a\\ \therefore\; \frac{{dy}}{{dx}} &= \frac{{ - a}}{{2by + \sin y}}\end{align}

## Chapter 5 Ex.5.3 Question 4

Find $$\frac{{dy}}{{dx}}\;\;:\;\;xy + {y^2} = \tan x + y$$

### Solution

Given, $$xy + {y^2} = \tan x + y$$

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( {xy + {y^2}} \right) &= \frac{d}{{dx}}\left( {\tan x + y} \right)\\ \Rightarrow\; \frac{d}{{dx}}\left( {xy} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) &= \frac{d}{{dx}}\left( {\tan x} \right) + \frac{{dy}}{{dx}}\\ \Rightarrow\; \left[ {y.\frac{d}{{dx}}\left( x \right) + x.\frac{{dy}}{{dx}}} \right] + 2y\frac{{dy}}{{dx}} &= {\sec ^2}x + \frac{{dy}}{{dx}} \qquad\qquad \left[ {{\text{using product rule and chain rule}}} \right]\\ \Rightarrow\; y.1 + x\frac{{dy}}{{dx}} + 2y\frac{{dy}}{{dx}} &= {\sec ^2}x + \frac{{dy}}{{dx}} \\ \Rightarrow\; \left( {x + 2y - 1} \right)\frac{{dy}}{{dx}} &= {\sec ^2}x - y\\ \therefore \;\frac{{dy}}{{dx}} &= \frac{{{{\sec }^2}x - y}}{{\left( {x + 2y - 1} \right)}}\end{align}

## Chapter 5 Ex.5.3 Question 5

Find $$\frac{{dy}}{{dx}} \;\; : \;\; {x^2} + xy + {y^2} = 100$$

### Solution

Given, $${x^2} + xy + {y^2} = 100$$

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( {{x^2} + xy + {y^2}} \right) &= \frac{d}{{dx}}\left( {100} \right)\\ \Rightarrow\; \frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( {xy} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) &= 0\\ \Rightarrow\; 2x + \left[ {y.\frac{d}{{dx}}\left( x \right) + x.\frac{{dy}}{{dx}}} \right] + 2y\frac{{dy}}{{dx}} &= 0\\ \Rightarrow\; 2x + y.1 + x.\frac{{dy}}{{dx}} + 2y\frac{{dy}}{{dx}} &= 0\\ \Rightarrow\; 2x + y + \left( {x + 2y} \right)\frac{{dy}}{{dx}}&= 0\\ \therefore \, \frac{{dy}}{{dx}} &= - \frac{{2x + y}}{{x + 2y}}\end{align}

## Chapter 5 Ex.5.3 Question 6

Find $$\frac{{dy}}{{dx}} \;\;: \;\;{x^3} + {x^2}y + x{y^2} + {y^3} = 81$$

### Solution

Given, $${x^3} + {x^2}y + x{y^2} + {y^3} = 81$$

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( {{x^3} + {x^2}y + x{y^2} + {y^3}} \right) &= \frac{d}{{dx}}\left( {81} \right)\\ \Rightarrow\; \frac{d}{{dx}}\left( {{x^3}} \right) + \frac{d}{{dx}}\left( {{x^2}y} \right) + \frac{d}{{dx}}\left( {x{y^2}} \right) + \frac{d}{{dx}}\left( {{y^3}} \right) &= 0\\ \Rightarrow\; 3{x^2} + \left[ {y\frac{d}{{dx}}\left( {{x^2}} \right) + {x^2}\frac{{dy}}{{dx}}} \right] + \left[ {{y^2}\frac{d}{{dx}}\left( x \right) + x\frac{d}{{dx}}\left( {{y^2}} \right)} \right] + 3{y^2}\frac{{dy}}{{dx}} &= 0\\ \Rightarrow\; 3{x^2} + \left[ {y.2x + {x^2}\frac{{dy}}{{dx}}} \right] + \left[ {{y^2}.1 + x.2y.\frac{{dy}}{{dx}}} \right] + 3{y^2}\frac{{dx}}{{dy}} &= 0\\ \Rightarrow\; \left( {{x^2} + 2xy + 3{y^2}} \right)\frac{{dy}}{{dx}} + \left( {3{x^2} + 2xy + {y^2}} \right) &= 0\\ \therefore\; \frac{{dy}}{{dx}} &= \frac{{ - \left( {3{x^2} + 2xy + {y^2}} \right)}}{{\left( {{x^2} + 2xy + 3{y^2}} \right)}}\end{align}

## Chapter 5 Ex.5.3 Question 7

Find $$\frac{{dx}}{{dy}}\;\;: \;\;{\sin ^2}y + \cos xy = \pi$$

### Solution

Given, $${\sin ^2}y + \cos xy = \pi$$

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( {{{\sin }^2}y + \cos xy} \right) &= \frac{d}{{dx}}\left( \pi \right)\\ \Rightarrow\; \frac{d}{{dx}}\left( {{{\sin }^2}y} \right) + \frac{d}{{dx}}\left( {\cos xy} \right) &= 0 \qquad\qquad\qquad\quad\;\;\;\; \ldots \left( 1 \right)\end{align}

Using chain rule, we obtain

\begin{align}\frac{d}{{dx}}\left( {{{\sin }^2}y} \right) &= 2\sin y\frac{d}{{dx}}\left( {\sin y} \right) \\&= 2\sin y\cos y\frac{{dy}}{{dx}} \qquad\qquad\qquad\qquad\qquad \ldots \left( 2 \right)\\[5pt] \frac{d}{{dx}}\left( {\cos xy} \right) &= - \sin xy\frac{d}{{dx}}\left( {xy} \right) \\&= - \sin xy\left[ {y\frac{d}{{dx}}\left( x \right) + x\frac{{dy}}{{dx}}} \right]\\&= - \sin xy\left[ {y.1 + x\frac{{dy}}{{dx}}} \right] \\&= - y\sin xy - x\sin xy\frac{{dy}}{{dx}} \qquad\qquad\qquad\;\;\; \ldots \left( 3 \right)\end{align}

From (1), (2) and (3), we obtain

\begin{align} 2\sin y\cos y\frac{{dy}}{{dx}} + \left( { - y\sin xy - x\sin xy\frac{{dy}}{{dx}}} \right) &= 0\\ \Rightarrow\; \left( {2\sin y\cos y - x\sin xy} \right)\frac{{dy}}{{dx}} &= y\sin xy\\ \Rightarrow\; \left( {\sin 2y - x\sin xy} \right)\frac{{dx}}{{dy}} &= y\sin xy\\ \therefore \; \frac{{dx}}{{dy}} &= \frac{{y\sin xy}}{{\sin 2y - x\sin xy}}\end{align}

## Chapter 5 Ex.5.3 Question 8

Find $$\frac{{dy}}{{dx}}\;\;:\;\;{\sin ^2}x + {\cos ^2}y = 1$$

### Solution

Given, $${\sin ^2}x + {\cos ^2}y = 1$$

Differentiating with respect to $$x$$, we get

\begin{align}\frac{d}{{dx}}\left( {{{\sin }^2}x + {{\cos }^2}y} \right) &= \frac{d}{{dx}}\left( 1 \right)\\ \Rightarrow\; \frac{d}{{dx}}\left( {{{\sin }^2}x} \right) + \frac{d}{{dx}}\left( {{{\cos }^2}y} \right) &= 0\\ \Rightarrow\; 2\sin x.\frac{d}{{dx}}\left( {\sin x} \right) + 2\cos y.\frac{d}{{dx}}\left( {\cos y} \right) &= 0\\ \Rightarrow\; 2\sin x\cos x + 2\cos y\left( { - \sin y} \right).\frac{{dy}}{{dx}} &= 0\\ \Rightarrow\; \sin 2x - \sin 2y\frac{{dy}}{{dx}} &= 0\\ \therefore \; \frac{{dy}}{{dx}} &= \frac{{\sin 2x}}{{\sin 2y}}\end{align}

## Chapter 5 Ex.5.3 Question 9

Find $$\frac{{dy}}{{dx}}\;\;:\;\;y = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$$

### Solution

Given,

\begin{align}y &= {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\\ \Rightarrow\; \; \sin y &= \frac{{2x}}{{1 + {x^2}}}\end{align}

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( {\sin y} \right) &= \frac{d}{{dx}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) \qquad \ldots \left( 1 \right)\\[5pt] \Rightarrow\; \; \cos y\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\end{align}

The function $$\frac{{2x}}{{1 + {x^2}}}$$, is of the form of $$\frac{u}{v}$$

By quotient rule, we get

\begin{align}\frac{d}{{dx}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) &= \frac{{\left( {1 + {x^2}} \right)\frac{d}{{dx}}\left( {2x} \right) - 2x.\frac{d}{{dx}}\left( {1 + {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\[5pt]&= \frac{{\left( {1 + {x^2}} \right).2 - 2x.\left[ {0 + 2x} \right]}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\[5pt]&= \frac{{2 + 2{x^2} - 4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\[5pt]&= \frac{{2\left( {1 - {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\end{align}

Also, $$\sin y = \frac{{2x}}{{1 + {x^2}}}$$

\begin{align}\cos y &= \sqrt {1 - {{\sin }^2}y} = \sqrt {1 - {{\left( {\frac{{2x}}{{1 + {x^2}}}} \right)}^2}} \\[5pt]&= \sqrt {\frac{{{{\left( {1 + {x^2}} \right)}^2} - 4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \\[5pt]&= \sqrt {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \\[5pt]&= \frac{{1 - {x^2}}}{{1 + {x^2}}}\end{align}

From ($$1$$), ($$2$$) and ($$3$$), we get

\begin{align}\frac{{1 - {x^2}}}{{1 + {x^2}}} \times \frac{{dy}}{{dx}} &= \frac{{2\left( {1 - {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ \Rightarrow\; \; \frac{{dy}}{{dx}} &= \frac{2}{{1 + {x^2}}}\end{align}

## Chapter 5 Ex.5.3 Question 10

Find $$\frac{{dy}}{{dx}}\;\;:\;\;y = {\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right), - \frac{1}{{\sqrt 3 }} < x < \frac{1}{{\sqrt 3 }}$$

### Solution

Given, $$y = {\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right)$$

$$\Rightarrow\; \; \tan y = \left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Since, we know that

$$\Rightarrow\; \; \tan y = \left( {\frac{{3\tan \frac{y}{3} - {{\tan }^3}\frac{y}{3}}}{{1 - 3{{\tan }^2}\frac{y}{3}}}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\left( 2 \right)$$

Comparing (1) and (2) we get,

$$x = \tan \frac{y}{3}$$

Differentiating with respect to $$x$$, we get

\begin{align}&\frac{d}{{dx}}\left( x \right) = \frac{d}{{dx}}\left( {\tan \frac{y}{3}} \right)\\ & \Rightarrow\; \; 1 = {\sec ^2}\frac{y}{3}.\frac{d}{{dx}}\left( {\frac{y}{3}} \right)\\ & \Rightarrow\; \; 1 = {\sec ^2}\frac{y}{3}.\frac{1}{3}.\frac{{dy}}{{dx}}\\ &\Rightarrow\; \; \frac{{dy}}{{dx}} = \frac{3}{{{{\sec }^2}\frac{y}{3}}} = \frac{3}{{1 + {{\tan }^2}\frac{y}{3}}}\\&\therefore\; \frac{{dy}}{{dx}} = \frac{3}{{1 + {x^2}}}\end{align}

## Chapter 5 Ex.5.3 Question 11

Find $$\frac{{dy}}{{dx}}\;\;:\;\;y = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) \,,0 < x < 1$$

### Solution

Given, $$y = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$$

\begin{align}& \Rightarrow\; \; \cos y = \left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\\ & \Rightarrow\; \; \frac{{1 - {{\tan }^2}\frac{y}{2}}}{{1 + {{\tan }^2}\frac{y}{2}}} = \frac{{1 - {x^2}}}{{1 + {x^2}}}\end{align}

Comparing LHS and RHS, we get

$$\tan \frac{y}{2} = x$$

Differentiating with respect to $$x$$, we get

\begin{align}&{\sec ^2}\frac{y}{2}.\frac{d}{{dx}}\left( {\frac{y}{2}} \right) = \frac{d}{{dx}}\left( x \right)\\ &\Rightarrow\; \; {\sec ^2}\frac{y}{2} \times \frac{1}{2}\frac{{dy}}{{dx}} = 1\\& \Rightarrow\; \; \frac{{dy}}{{dx}} = \frac{2}{{{{\sec }^2}\frac{y}{2}}}\\ & \Rightarrow\; \; \frac{{dy}}{{dx}} = \frac{2}{{1 + {{\tan }^2}\frac{y}{2}}}\\&\therefore \;\frac{{dy}}{{dx}} = \frac{2}{{1 + {x^2}}}\end{align}

## Chapter 5 Ex.5.3 Question 12

Find $$\frac{{dy}}{{dx}}\;\;:\;\;y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1$$

### Solution

Given, $$y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$$

$$y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$$

$$\Rightarrow\; \; \sin y = \frac{{1 - {x^2}}}{{1 + {x^2}}}$$

Differentiating with respect to $$x$$, we get

$$\frac{d}{{dx}}\left( {\sin y} \right) = \frac{d}{{dx}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Using chain rule, we get

$$\frac{d}{{dx}}\left( {\sin y} \right) = \cos y.\frac{{dy}}{{dx}}$$

\begin{align}\cos y &= \sqrt {1 - {{\sin }^2}y} = \sqrt {1 - {{\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)}^2}} \\ &= \sqrt {\frac{{{{\left( {1 + {x^2}} \right)}^2} - {{\left( {1 - {x^2}} \right)}^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}}\\& = \sqrt {\frac{{4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \\&= \frac{{2x}}{{1 + {x^2}}}\end{align}

Therefore,

$$\frac{d}{{dx}}\left( {\sin y} \right) = \frac{{2x}}{{1 + {x^2}}}\frac{{dy}}{{dx}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)$$

\begin{align}\frac{d}{{dx}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) &= \frac{{\left( {1 + {x^2}} \right).\frac{d}{{dx}}\left( {1 - {x^2}} \right) - \left( {1 - {x^2}} \right).\frac{d}{{dx}}\left( {1 + {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} \qquad \left[ {{\text{using quotient rule}}} \right]\\ &= \frac{{\left( {1 + {x^2}} \right)\left( { - 2x} \right) - \left( {1 - {x^2}} \right)\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ &= \frac{{ - 2x - 2{x^3} - 2x + 2{x^3}}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ &= \frac{{ - 4x}}{{{{\left( {1 + {x^2}} \right)}^2}}} \qquad \ldots \left( 3 \right)\end{align}

From equation (1), (2) and (3), we get

\begin{align}&\frac{{2x}}{{1 + {x^2}}}\frac{{dy}}{{dx}} = \frac{{ - 4x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ &\Rightarrow\; \; \frac{{dy}}{{dx}} = \frac{{ - 2}}{{1 + {x^2}}}\end{align}

## Chapter 5 Ex.5.3 Question 13

Find $$\frac{{dy}}{{dx}}\;\;:\;\;y = {\cos ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right), - 1 < x < 1$$

### Solution

Given, $$y = {\cos ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$$

$$y = {\cos ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$$

$$\cos y = \left( {\frac{{2x}}{{1 + {x^2}}}} \right)$$

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( {\cos y} \right) &= \frac{d}{{dx}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\\ \Rightarrow\; \; - \sin y.\frac{{dy}}{{dx}} &= \frac{{\left( {1 + {x^2}} \right).\frac{d}{{dx}}\left( {2x} \right) - 2x.\frac{d}{{dx}}\left( {1 + {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ \Rightarrow\; \; - \sqrt {1 - {{\cos }^2}y} \frac{{dy}}{{dx}} &= \frac{{\left( {1 + {x^2}} \right) \times 2 - 2x \times 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ \Rightarrow\; \; \left[ {\sqrt {1 - {{\left( {\frac{{2x}}{{1 + {x^2}}}} \right)}^2}} } \right]\frac{{dy}}{{dx}} &= - \left[ {\frac{{2\left( {1 - {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \right]\\ \Rightarrow\; \; \sqrt {\frac{{{{\left( {1 + {x^2}} \right)}^2} - 4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}} .\frac{{dy}}{{dx}} &= \frac{{ - 2\left( {1 - {x^2}} \right)}}{{\left( {1 + {x^2}} \right)}}\\ \Rightarrow\; \; \sqrt {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \frac{{dy}}{{dx}} &= \frac{{ - 2\left( {1 - {x^2}} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\\ \Rightarrow\; \; \frac{{1 - {x^2}}}{{1 + {x^2}}}.\frac{{dy}}{{dx}} &= \frac{{ - 2\left( {1 - {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\ \Rightarrow\; \; \frac{{dy}}{{dx}} &= \frac{{ - 2}}{{1 + {x^2}}}\end{align}

## Chapter 5 Ex.5.3 Question 14

Find $$\frac{{dy}}{{dx}}\;\;:\;\;y = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right), - \frac{1}{{\sqrt 2 }} < x < \frac{1}{{\sqrt 2 }}$$

### Solution

Given, $$y = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$$

$$y = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$$

$$\Rightarrow\; \; \sin y = \left( {2x\sqrt {1 - {x^2}} } \right)$$

Differentiating with respect to $$x$$, we get

\begin{align} \cos y.\frac{{dy}}{{dx}} &= 2\left[ {x\frac{d}{{dx}}\left( {\sqrt {1 - {x^2}} } \right) + \sqrt {1 - {x^2}} \frac{{dx}}{{dx}}} \right]\\ \Rightarrow\; \; \sqrt {1 - {{\sin }^2}y} \frac{{dy}}{{dx}} &= 2\left[ {\frac{x}{2}.\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }} + \sqrt {1 - {x^2}} } \right]\\ \Rightarrow\; \; \sqrt {1 - {{\left( {2x\sqrt {1 - {x^2}} } \right)}^2}} .\frac{{dy}}{{dx}} &= 2\left[ {\frac{{ - {x^2} + 1 - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right]\\ \Rightarrow\; \; \sqrt {1 - 4{x^2}\left( {1 - {x^2}} \right)} \frac{{dy}}{{dx}} &= 2\left[ {\frac{{1 - 2{x^2}}}{{\sqrt {1 - {x^2}} }}} \right]\\ \Rightarrow\; \; \sqrt {{{\left( {1 - 2{x^2}} \right)}^2}} \frac{{dy}}{{dx}} &= 2\left[ {\frac{{1 - 2{x^2}}}{{\sqrt {1 - {x^2}} }}} \right]\\ \Rightarrow\; \; \left( {1 - 2{x^2}} \right)\frac{{dy}}{{dx}} &= 2\left[ {\frac{{1 - 2{x^2}}}{{\sqrt {1 - {x^2}} }}} \right]\\ \Rightarrow\; \; \frac{{dy}}{{dx}} &= \frac{2}{{\sqrt {1 - {x^2}} }}\end{align}

## Chapter 5 Ex.5.3 Question 15

Find $$\frac{{dy}}{{dx}}\;\;:\;\;y = {\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} \right),\;0 < x < \frac{1}{{\sqrt 2 }}$$

### Solution

Given, $$y = {\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} \right)$$

\begin{align} \Rightarrow\; \; y &= {\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} \right)\\ \Rightarrow\; \; \sec y &= \left( {\frac{1}{{2{x^2} - 1}}} \right)\\ \Rightarrow\; \; \cos y &= 2{x^2} - 1\\ \Rightarrow\; \; 2{x^2} &= 1 + \cos y\\ \Rightarrow\; \; 2{x^2} &= 2{\cos ^2}\frac{y}{2}\\ \Rightarrow\; \; x &= \cos \frac{y}{2}\end{align}

Differentiating with respect to $$x$$, we get

\begin{align} \frac{d}{{dx}}\left( x \right) &= \frac{d}{{dx}}\left( {\cos \frac{y}{2}} \right)\\ \Rightarrow\; \; 1 &= \sin \frac{y}{2}.\frac{d}{{dx}}\left( {\frac{y}{2}} \right)\\ \Rightarrow\; \; \frac{{ - 1}}{{\sin \frac{y}{2}}} &= \frac{1}{2}\frac{{dy}}{{dx}}\\ \Rightarrow\; \; \frac{{dy}}{{dx}} &= \frac{{ - 2}}{{\sin \frac{y}{2}}}\\ \Rightarrow\; \; \frac{{dy}}{{dx}} &= \frac{{ - 2}}{{\sqrt {1 - {{\cos }^2}\frac{y}{2}} }}\\ \Rightarrow\; \; \frac{{dy}}{{dx}} &= \frac{{ - 2}}{{\sqrt {1 - {x^2}} }}\end{align}

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