Chapter 5: Continuity and Differentiability

Ans: The given relationship is,

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

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

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

On comparing L.H.S. and R.H.S. of the above relationship, 

We get 

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

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

Differentiating the above relationship with respect to

x, 

We get

$ \Rightarrow {\text{ }}$   $\frac{{dy}}{{dx}} = \frac{{d\left( {2{{\tan }^{ - 1}}x} \right)}}{{dx}}$ 

$ \Rightarrow \frac{{dy}}{{dx}} = 2\frac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}}$ 

$ \Rightarrow \frac{{dy}}{{dx}} = 2\left( {\frac{1}{{1 + {x^2}}}} \right)$ 

$\therefore \frac{{dy}}{{dx}} = \frac{2}{{1 + {x^2}}}$

Ans: The given relationship is  $ 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}}} $

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

$ \Rightarrow (1+\sin y){{x}^{2}}=1-\sin y $

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

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

$ \Rightarrow x=\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}} $

$ \Rightarrow x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}} $

$ \Rightarrow x=\tan \left( \frac{\pi }{4}-\frac{\pi }{2} \right) $

Differentiating the above relationship with respect to $ x $ , 

We get,

$ \Rightarrow \text{  } $   $ \frac{d}{dx}(x)=\frac{d}{dx}\left[ \tan \left( \frac{\pi }{4}-\frac{y}{2} \right) \right] $

$ \Rightarrow 1={{\sec }^{2}}\left( \frac{\pi }{4}-\frac{y}{2} \right)\cdot \frac{d}{dt}\left( \frac{\pi }{4}-\frac{y}{2} \right) $

$ \Rightarrow 1=\left[ 1+{{\tan }^{2}}\left( \frac{\pi }{4}-\frac{y}{2} \right)\left( -\frac{1}{2}\frac{dy}{dx} \right) \right. $

$ \Rightarrow 1=\left( 1+{{x}^{2}} \right)\left( -\frac{1}{2}\frac{dy}{dx} \right) $

$ \Rightarrow \frac{dy}{dx}=\frac{-2}{1+{{x}^{2}}} $

Ans: The given relationship is 

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

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

$ \Rightarrow \cos y=\frac{2x}{1+{{x}^{2}}} $

Differentiating the above relationship with respect to $ \text{x} $ , 

We get,

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

$ \Rightarrow -\sin y\frac{dy}{dx}=\frac{\left( 1+{{x}^{2}} \right)\cdot \frac{d}{dx}(2x)-2x\cdot \frac{d}{dt}\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\cdot 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{{(1-{{x}^{2}})}}}{{{\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)}^{2}}} $

$ =\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}}}\cdot \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}}} $

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

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

$ \Rightarrow \sin y=2x\sqrt{1-{{x}^{2}}} $

Differentiating the above relationship with respect to  $ x $ , we get  $ \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] $

$ =\sqrt{1-{{\sin }^{2}}y}\frac{dy}{dx}=2\left[ \frac{x}{2}\cdot \frac{-2x}{\sqrt{1-{{x}^{2}}}}+\sqrt{1-{{x}^{2}}} \right] $

$ =\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] $

$ =\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] $

$ =\sqrt{{{(1-2{x}^{2})}^{2}}}\frac{dy}{dx}=2\left[ \frac{1-2{{x}^{2}}}{\sqrt{1-{{x}^{2}}}} \right] $

$ =\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}}}} $

Ans:  The given relationship is  $ y={{\sec }^{-1}}\left( \frac{1}{2{{x}^{2}}-1} \right) $ 

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

$ \Rightarrow \sec y=\frac{1}{2{{x}^{2}}-1} $

$ \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} $

Differentiating the above relationship with respect to $ x $ , 

We get,

$ \Rightarrow \text{  } $  $ \frac{d}{dx}(x)=\frac{d}{dx}\left( \cos \frac{y}{2} \right) $

$ \Rightarrow 1=-\sin \frac{y}{2}\cdot \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}}=\frac{-2}{\sqrt{1-{{\cos }^{2}}\frac{y}{2}}} $

$ \Rightarrow \frac{dy}{dx}=\frac{-2}{\sqrt{1-{{x}^{2}}}} $