Ncert Solutions For Class 12 Maths Ex 5.3

Ncert Solutions For Class 12 Maths Chapter 5 Ex 5.3

Q1: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(2x+3y=\sin x\)

Sol:

The given relationship is \(2x+3y=\sin x\)

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(2x+3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)\) \(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (2x) + \frac{\mathrm{d} }{\mathrm{d} x}(3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)\) \(\Rightarrow 2 + 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos x\) \(\Rightarrow 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos x-2\)

\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\cos x-2}{3}\).

Q2: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(2x+3y=\sin y\).

Sol:

The given relationship is \(2x+3y=\sin y\)

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(2x+3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)\) \(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (2x) + \frac{\mathrm{d} }{\mathrm{d} x}(3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)\) \(\Rightarrow 2 + 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos y \frac{\mathrm{d} y}{\mathrm{d} x}\) \(\Rightarrow 2=(\cos y-3) \frac{\mathrm{d} y}{\mathrm{d} x}\)

\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\cos y -3}\).

Q3: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(ax+by^{2}=\cos y\).

Sol:

The given relationship is \(ax+by^{2}=\cos y\)

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(ax+by^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)\) \(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (ax) + \frac{\mathrm{d} }{\mathrm{d} x}(by^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)\) \(\Rightarrow a + 2by\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin y \frac{\mathrm{d} y}{\mathrm{d} x}\) \(\Rightarrow a=-(\sin y+2by) \frac{\mathrm{d} y}{\mathrm{d} x}\)

\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-a}{(\sin y+2by )}\).

Q4: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(xy+y^{2}=\tan x+ y\).

Sol:

The given relationship is \(xy+y^{2}=\tan x+ y\)

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(xy+y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\tan x + y)\) \(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (xy) + \frac{\mathrm{d} }{\mathrm{d} x}(y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\tan x)+ \frac{\mathrm{d} y}{\mathrm{d} x} \) \(\Rightarrow y.\frac{\mathrm{d} }{\mathrm{d} x} (x)+x .\frac{\mathrm{d} y}{\mathrm{d} x} +2y. \frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x+ \frac{\mathrm{d} y}{\mathrm{d} x} \) \(\Rightarrow y.1 + x.\frac{\mathrm{d} y}{\mathrm{d} x}+2y. \frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x+ \frac{\mathrm{d} y}{\mathrm{d} x}\) \(\Rightarrow (x+2y-1)\frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x-y\)

\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sec ^{2}x-y}{(x+2y-1 )}\).

Q5: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(x^{2}+xy+y^{2}=100\).

Sol:

The given relationship is \(x^{2}+xy+y^{2}=100\)

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}+xy+y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(100)\)

\(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (x^{2})+\frac{\mathrm{d} }{\mathrm{d} x} (xy) + \frac{\mathrm{d} }{\mathrm{d} x}(y^{2})=0 \) (derivatives of constant function is 0)

\(\Rightarrow 2x+ y.1+x.\frac{\mathrm{d} y}{\mathrm{d} x} +2y.\frac{\mathrm{d} y}{\mathrm{d} x}=0 \) \(\Rightarrow 2x+ y+(x+2y)\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{2x+y}{x+2y}\)

 

Q6: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(x^{3}+x^{2}y+xy^{2}+y^{3}\).

Sol:

The given relationship is \(x^{3}+x^{2}y+xy^{2}+y^{3}=81\)

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(x^{3}+x^{2}y+xy^{2}+y^{3})=\frac{\mathrm{d} }{\mathrm{d} x}(81)\)

\(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (x^{3})+\frac{\mathrm{d} }{\mathrm{d} x} (x^{2}y) + \frac{\mathrm{d} }{\mathrm{d} x}(xy^{2})+\frac{\mathrm{d} }{\mathrm{d} x}(y^{3})=0 \) (derivatives of constant function is 0)

\(\Rightarrow 3x^{2}+ y.\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})+x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+y^{2}\frac{\mathrm{d} }{\mathrm{d} x}(x)+x\frac{\mathrm{d} }{\mathrm{d} x}(y^{2}+3y^{2}.\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(\Rightarrow 3x^{2}+ y.2x+x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+y^{2}.1+x.2y.\frac{\mathrm{d} y}{\mathrm{d} x}+3y^{2}\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(\Rightarrow (x^{2}+2xy+3y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+(3x^{2}+2xy+y^{2}=0\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-(3x^{2}+2xy+y^{2})}{x^{2}+2xy+3y^{2}}\)

 

Q7: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(\sin ^{2}y +\cos xy=\Pi\).

Sol:

The given relationship is \(\sin ^{2}y+\cos xy=\Pi\)

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}y+\cos xy)=\frac{\mathrm{d} }{\mathrm{d} x}(\Pi)\)

\(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (\sin ^{2}y)+\frac{\mathrm{d} }{\mathrm{d} x} (\cos xy) =0 \) ……(i) (derivatives of constant function is 0)

Using chain rule,we get

\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}y=2 \sin y \frac{\mathrm{d} }{\mathrm{d} x}(\sin y) =2\sin y \cos y \frac{\mathrm{d} y}{\mathrm{d} x}\) …….(ii)

\(\frac{\mathrm{d} }{\mathrm{d} x}(\cos xy)=-\sin xy \frac{\mathrm{d} }{\mathrm{d} x}(xy)= -\sin xy (y. \frac{\mathrm{d} }{\mathrm{d} x}(x)+x. \frac{\mathrm{d} y}{\mathrm{d} x})= -\sin xy (y. 1+x. \frac{\mathrm{d} y}{\mathrm{d} x})= -y \sin xy -x\sin xy \frac{\mathrm{d} y}{\mathrm{d} x}\) …….(iii)

From (i), (ii) and (iii) we have

\( 2\sin y \cos y \frac{\mathrm{d} y}{\mathrm{d} x}- y\sin xy- x \sin xy \frac{\mathrm{d} y}{\mathrm{d} x}=0\) \(\Rightarrow (2\sin y \cos y- x \sin xy)\frac{\mathrm{d} y}{\mathrm{d} x}= y \sin xy\) \(\Rightarrow (\sin 2y-x\sin xy )\frac{\mathrm{d} y}{\mathrm{d} x}=y \sin xy\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{y \sin xy}{\sin 2y -x \sin xy}\)

 

Q8: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(\sin ^{2}x +\cos ^{y}=1\).

Sol:

The given relationship is \(\sin ^{2}x +\cos ^{y}=1\)

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}x+\cos ^{y})=\frac{\mathrm{d} }{\mathrm{d} x}(1)\)

\(\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (\sin ^{2}y)+\frac{\mathrm{d} }{\mathrm{d} x} (\cos ^{y}) =0 \) (derivatives of constant function is 0)

\(\Rightarrow 2\sin x \frac{\mathrm{d} }{\mathrm{d} x} (\sin x)+2\cos y.\frac{\mathrm{d} }{\mathrm{d} x} (\cos y) =0 \) \(\Rightarrow 2\sin x \cos x+2\cos y(-\sin y).\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(\Rightarrow \sin 2x-\sin 2y.\frac{\mathrm{d} y}{\mathrm{d} x} =0 \) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin 2x}{\sin 2y}\)

 

Q9: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

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

Sol:

The given relationship is \(y=\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right )\)

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

Differentiating the equation with respect to x, we have

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

\(\Rightarrow \cos y \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{2x}{1+x^{2}} \right ) \) ………..(i)

The right side function \( \left ( \frac{2x}{1+x^{2}} \right ) \) is of the form \(\frac{u}{v}\)

So, by quotient rule, we obtain

\(\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )=\frac{(1+x^{2}).\frac{\mathrm{d} }{\mathrm{d} x}(2x)-2x.\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})}{(1+x^{2})^{2}}\)

\(=\frac{(1+x^{2}).2-2x.(0+2x)}{(1+x^{2})^{2}}=\frac{2+2x^{2}-4x^{2}}{(1+x^{2})^{2}}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}\) ………(ii)

Also \(\sin y=\frac{2x}{1+x^{2}}\)

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

\(=\sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}=\frac{1-x^{2}}{1+x^{2}}\) ……….(iii)

From (i) , (ii) and (iii) we obtain

\(=\frac{1-x^{2}}{1+x^{2}}\times \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+x^{2}}\)

 

Q10: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(y=\tan ^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right ),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\).

Sol:

The given relationship is \(y=\tan ^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )\)

\(\tan y=\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )\) ………(i)

Differentiating the equation with respect to x, we have

We know that, \(\tan y=\frac{3\tan \frac{y}{3}-\tan ^{3}\frac{y}{3}}{1-3\tan ^{2}\frac{y}{3}}\) …………(ii)

Comparing equation (i) and (ii), we have

\(x=\tan \frac{y}{3}\)

Differentiating this relationship w.r.t. x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(x)=\frac{\mathrm{d} }{\mathrm{d} x} \left ( \tan \frac{y}{3} \right )\) \(\Rightarrow 1=\sec ^{2}\frac{y}{3}.\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{y}{3} \right )\) \(\Rightarrow 1=\sec ^{2}\frac{y}{3}.\frac{1}{3}\frac{\mathrm{d} y}{\mathrm{d} x} \) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{\sec ^{2}\frac{y}{3}}=\frac{3}{1+\tan ^{2}\frac{y}{3}}\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{1+x^{2}}\)

 

Q11: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(y=\cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1\).

Sol:

The given relationship is \(y=\cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1\)

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

Comparing both sides equation

\(\tan \frac{y}{2}=x\)

Differentiating the equation with respect to x, we have

\(\sec ^{2} \frac{y}{2}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{2} \right )=\frac{\mathrm{d} }{\mathrm{d} x}(x)\) \(\sec ^{2} \frac{y}{2}\times \frac{1}{2}.\frac{\mathrm{d} y}{\mathrm{d} x}=1\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\sec ^{2}\frac{y}{2}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+\tan ^{2}\frac{y}{2}}\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+x ^{2}}\)

 

Q12: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(y=\sin ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1\).

Sol:

The given relationship is \(y=\sin ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1\)

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

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )\) ……..(i)

Using chain rule

\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\cos y \frac{\mathrm{d}y }{\mathrm{d} x}\) \(\cos y=\sqrt{1-\sin ^{2}y}=\sqrt{1-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}\) \(=\sqrt{({1+x^{2}})^{2}-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}\) \(=\sqrt{ \frac{({1+x^{2}})^{2}-(1-x^{2})}{(1+x^{2})^{2}} ^{2}}=\sqrt{\frac{4x^{2}}{(1+x^{2})^{2}}}=\frac{2x}{1+x^{2}}\)

\(therefore,\; \frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{2x}{1+x^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}\) …..(ii)

\(\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )=\frac{(1+x^{2}).(1-x^{2})-(1-x^{2}).(1+x^{2})}{(1+x^{2})^{2}}\) [using quotient rule]

\(=\frac{(1+x^{2}).(-2x)-(1-x^{2}).(2x)}{(1+x^{2})^{2}}\) \(=\frac{-2x-2x^{3}-2x+2x^{3}}{(1+x^{2})^{2}}\)

\(=\frac{-4x}{(1+x^{2})^{2}}\) ……..(iii)

From (i),(ii) and (iii) we have

\(\frac{2x}{(1+x^{2})}\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-4x}{(1+x^{2})^{2}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{(1+x^{2})}\)

 

Q13: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

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

Sol:

The given relationship is \(y=\cos ^{-1}\left ( \frac{2x}{1+x^{2}} \right )\)

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

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )\) \(-\sin y \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{(1+x^{2}.\frac{\mathrm{d} }{\mathrm{d} x}(2x)-2x.\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})}{(1+x^{2})^{2}}\) \(\Rightarrow -\sqrt{1-\cos^{2}y}.\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{(1+x^{2}.2-2x.2x)}{(1+x^{2})^{2}}\) \(\Rightarrow \sqrt{1-\left ( \frac{2x}{1+x^{2}} \right )^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]\) \(\Rightarrow \sqrt{\frac{(1+x^{2})^{2}-4x^{2}}{(1+x^{2})^{2}}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]\) \(\Rightarrow \sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]\) \(\Rightarrow \frac{(1-x^{2})}{(1+x^{2})}.\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{1+x^{2}}\)

 

Q14: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

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

Sol:

The given relationship is \(y=\sin^{-1}(2x\sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\)

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

Differentiating the equation with respect to x, we have

\(\cos y{\mathrm{d} y}{\mathrm{d} x}=2\left [ x.\frac{\mathrm{d} }{\mathrm{d} x}\sqrt{1-x^{2}} +\sqrt{1-x^{2}}\frac{\mathrm{d} x}{\mathrm{d} x}\right ]\) \(\sqrt{1-\sin^{2}y} \frac{\mathrm{d}y }{\mathrm{d} x}=2\left [ \frac{x}{2}.\frac{-2x}{\sqrt{1-x^{2}}} +\sqrt{1-x^{2}}\right ]\) \(\Rightarrow \sqrt{1-(2x\sqrt{1-x^{2}})^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}} \right ]\) \(\Rightarrow \sqrt{1-4x^{2}(1-x^{2})}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{1-2x^{2}}{\sqrt{1-x^{2}}} \right ]\) \(\Rightarrow \sqrt{(1-2x^{2})^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{1-2x^{2}}{\sqrt{1-x^{2}}} \right ]\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\sqrt{1-x^{2}}}\)

 

Q15: Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

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

Sol:

The given relationship is \(y=\sec^{-1}\left ( \frac{1}{2x^{2}-1} \right ),0<x<\frac{1}{\sqrt{2}}\)

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

Differentiating the equation with respect to x, we have

\(\frac{\mathrm{d} }{\mathrm{d} x}(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos \frac{y}{2} \right )\) \(\Rightarrow 1=-\sin \frac{y}{2}.\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{2} \right )\) \(\Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2}\frac{\mathrm{d} y}{\mathrm{d} x}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos^{2}\frac{y}{2}}}\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{\sqrt{1-x^{2}}}\)