NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability - Exercise 5.3

Exercise 5.3

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

Ans: The given relationship is  $ 2x+3y=\sin x $  

Differentiating the above relationship with respect to $ x $ , 

We get

  $ \Rightarrow \text{  } $  $ \frac{d}{dy}(2x+3y)=\frac{d}{dx}(\sin x) $ 

 $ \Rightarrow     \frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\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} $ 

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

Ans: The given relationship is  $ 2x+3y=\sin y $  

Differentiating the above relationship with respect to $ x $ , We obtain

 $ \Rightarrow \text{  } $   $ \frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin y) $ 

 $ \Rightarrow 2+3\frac{dy}{dx}=\cos y\frac{dy}{dx}\quad [ $  By using chain rule]

 $ \Rightarrow 2=(cosy $  $ -3)\frac{dy}{dx} $ 

 $ \therefore \frac{dy}{dx}=\frac{2}{\cos y-3} $ 

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

Ans: The given relationship is $ ax+b{{y}^{2}}=\cos y $ .

Differentiating the above relationship with respect to $ x $ , we obtain

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

 $ \Rightarrow a+b\frac{d}{dx}\left( {{y}^{2}} \right)=\frac{d}{dx}(\cos y) $ 

 $ \frac{d}{dx}\left( {{y}^{2}} \right)=2y\frac{dy}{dx} $  and  $ \frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx} $ 

Using the chain rule, 

We get,

  $ \Rightarrow \text{  } $  $ a+b\times 2y\frac{dy}{dx}=-\sin y\frac{dy}{dx} $ 

 $ \Rightarrow (2by+\sin y)\frac{dy}{dx}=-a $ 

 $ \therefore \frac{dy}{dx}=\frac{-a}{2by+\sin y} $ 

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

Ans:  The given relationship is  $ xy+{{y}^{2}}=\tan x+y $ 

 Differentiating the above relationship with respect to $ x $ , We obtain

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

 $ \Rightarrow \frac{d}{dx}(xy)+\frac{d}{dx}\left( {{y}^{2}} \right)=\frac{d}{dx}(\tan x)+\frac{dy}{dx} $ 

 $ \Rightarrow \left[ y\cdot \frac{d}{dx}(x)+x\cdot \frac{dy}{dx} \right]+2y\frac{dy}{dx}={{\sec }^{2}}x+\frac{dy}{dx}\quad  $  [Using product rule and chain rule]

 $ \Rightarrow y\cdot 1+x\frac{dy}{dx}+2y\frac{dy}{dx}={{\sec }^{2}}x+\frac{dy}{dx} $ 

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

 $ \therefore \frac{dy}{dx}=\frac{{{\sec }^{2}}x-y}{(x+2y-1)} $ 

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

Ans:  The given relationship is $ {{x}^{2}}+xy+{{y}^{2}}=100 $ .

Differentiating the above relationship with respect to $ x $ , We obtain

 $ \Rightarrow \text{  } $   $ \frac{d}{dx}\left( {{x}^{2}}+xy+{{y}^{2}} \right)=\frac{d}{dx}(100) $                   Derivative of the constant function is  0 

 $ \Rightarrow \frac{d}{dx}\left( {{x}^{2}} \right)+\frac{d}{dx}(xy)+\frac{d}{dx}\left( {{y}^{2}} \right)=0 $ 

 $ \Rightarrow 2x+\left[ y\cdot \frac{d}{dx}(x)+x\cdot \frac{dy}{dx} \right]+2y\frac{dy}{dx}=0 $   Using product rule and chain rule

 $ \Rightarrow 2x+y.1+x\cdot \frac{dy}{dx}+2y\frac{dy}{dx}=0 $ 

 $ \Rightarrow 2x+y+(x+2y)\frac{dy}{dx}=0 $ 

 $ \therefore \frac{dy}{dx}=-\frac{2x+y}{x+2y} $ 

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

Ans: The given relationship is  $ {{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}}=81 $  .

Differentiating the above relationship with respect to  $ x $ , 

We get,

 $ \Rightarrow \text{  } $   $ \frac{d}{dx}\left( {{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}} \right)=\frac{d}{dx}(81) $ 

 $ \Rightarrow \frac{d}{dx}\left( {{x}^{3}} \right)+\frac{d}{dx}\left( {{x}^{2}}y \right)+\frac{d}{dx}{x{y}^{2}}+\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}(x)+x\frac{d}{dx}\left( {{y}^{2}} \right) \right]+3{{y}^{2}}\frac{dy}{dx}=0 $ 

 $ \Rightarrow 3{{x}^{2}}+\left[ y\cdot 2x+{{x}^{2}}\frac{dy}{dx} \right]+\left[ {{y}^{2}}\cdot 1+x\cdot 2y\cdot \frac{dy}{dx} \right]+3{{y}^{2}}\frac{dy}{dx}=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)} $ 

7. Find  $ \frac{dx}{dy}:{{\sin }^{2}}y+\cos xy=k $ 

Ans:   The given relationship is  $ {{\sin }^{2}}y+\cos xy=k  $ 

Differentiating the above relationship with respect to $ x $ , 

We get,

 $ \Rightarrow \text{  } $   $ \frac{d}{dx}\left( {{\sin }^{2}}y+\cos xy \right)=\frac{d}{dx}(k ) $ 

 $ \Rightarrow \frac{d}{dx}\left( {{\sin }^{2}}y \right)+\frac{d}{dx}(\cos xy)=0 $ 

Using the chain rule, 

We get  $ \frac{d}{dx}\left( {{\sin }^{2}}y \right)=2\sin y\frac{d}{dx}(\sin y)=2\sin y\cos y\frac{dy}{dx} $ 

 $ \frac{d}{dx}(\cos xy)=-\sin xy\frac{d}{dx}(xy)=-\sin xy\left[ y\frac{d}{dx}(x)+x\frac{dy}{dx} \right] $ 

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

From the above equations we get  $ 2\sin y\cos y\frac{dy}{dx}-y\sin xy-x\sin xy\frac{dy}{dx}=0 $ 

 $ \Rightarrow (2\sin y\cos y-x\sin xy)\frac{dy}{dx}=y\sin xy $ 

 $ \Rightarrow (\sin 2y-x\sin xy)\frac{dx}{dy}=y\sin xy $ 

 $ \frac{dx}{dy}=\frac{y\sin xy}{\sin 2y-x\sin xy} $ 

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

Ans:  The given relationship is  $ {{\sin }^{2}}x+{{\cos }^{2}}y=1 $ 

Differentiating the above relationship with respect to  $ x $ ,

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

 $ \Rightarrow \frac{d}{dx}\left( {{\sin }^{2}}x \right)+\frac{d}{dx}\left( {{\cos }^{2}}y \right)=0 $ 

 $ \Rightarrow 2\sin x\cdot \frac{d}{dx}(\sin x)+2\cos y\cdot \frac{d}{dx}(\cos y)=0 $ 

 $ \Rightarrow 2\sin x\cos x+2\cos y(-\sin y)\cdot \frac{dy}{dx}=0 $ 

 $ \sin 2x-\sin 2y\frac{dy}{dx}=0 $ 

 $ \frac{dx}{dy}=\frac{\sin 2x}{\sin 2y} $ 

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

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

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

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

Differentiating the above relationship with respect to  $ x $ , 

We get,

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

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

The function $ \frac{2x}{1+{{x}^{2}}} $  is of the form of $ \frac{u}{v} $ . Therefore, by quotient rule, we get  $ \frac{d}{dx}\left( \frac{2x}{1+{{x}^{2}}} \right)=\frac{\left( 1+{{x}^{2}} \right)\frac{d}{dx}(2x)-2x\cdot \frac{d}{dx}\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

 $ =\frac{\left( 1+{{x}^{2}} \right)\cdot 2-2x[0+2x]}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\frac{2+2{{x}^{2}}-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\frac{2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} $ 

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{{{\left( 1+{{x}^{2}} \right)}^{2}}-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}} $ 

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

From above equations , we get

 $ \Rightarrow \text{  } $   $ \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}}} $ 

10. Find $\dfrac{d y}{d x} \text { in, } y=\tan ^{-1}\left(\dfrac{3 x-x^{3}}{1-3 x^{2}}\right),-\dfrac{1}{\sqrt{3}} < x < \dfrac{1}{\sqrt{3}}$

Ans: $y=\tan ^{-1}\left(\dfrac{3 x-x^{3}}{1-3 x^{2}}\right)$

Putting $\mathrm{x}=\tan \theta$

$y=\tan ^{-1}\left(\dfrac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right) \\$

$y=\tan ^{-1}(\tan 3 \theta) \quad\left(\tan 3 \theta=\dfrac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right) \\$

$y=3 \theta$

Differentiating both sides w.r.t. $x$.

$\dfrac{d(y)}{d x}=\dfrac{d 3\left(\tan ^{-1} x\right)}{d x} \\$

$\dfrac{d y}{d x}=3 \dfrac{d\left(\tan ^{-1} x\right)}{d x} \\$

$\dfrac{d y}{d x}=3\left(\dfrac{1}{1+x^{2}}\right) \quad\left(\left(\tan ^{-1} x\right)^{\prime}=\dfrac{1}{1+x^{2}}\right) \\$

$\dfrac{d y}{d x}=\dfrac{3}{1+x^{2}}$

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

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 $ 

Differentiating the above relationship with respect to $ x $ , 

We get

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

 $ \Rightarrow {{\sec }^{2}}\frac{y}{2}\text{x}\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{1}{1+{{x}^{2}}} $ 

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

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}}} $ 

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

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}}} $ 

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

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

Important Points

• Defination of Continuity: A function f(x) is referred to as continuous at x=a; where a∈ domain of f(x), if

$\lim _{x \rightarrow a^{-1}} f(x)=\lim _{x \rightarrow a^{+1}} f(x)=f(a)$

i.e., Left Hand Limit = Right Hand Limit= Value of a function at x=a

or $\lim _{x \rightarrow a} f(x)=f(a)$

• Properties of Continous Fuctions are as follows.

If g and h are two real functions, continuous at real number c then,

(i) g + h is continuous at x = c.

(ii) g – h is continuous at x = c.

(iii) g.h is continuous at x = c.

(iv) cg is continuous, where c is any constant.

(v) (gh) is continuous at x = c, (provide g(c) ≠ 0)

• Definition of Differentianlity: A function f(x) is called differentiable at a point x = a, if LHD at (x = a) = RHD at (x = a)

i.e. Left hand derivative at (x = a) = Right hand derivative (at x = a)

Right Hand Derivative,$R f^{\prime}(a)=\lim _{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}$

Left Hand Derivative, $\operatorname{Lf}^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}$

• Rolle’s Theorem: Let f : [p, q] → R is continuous on [p, q] and differentiable on (p, q) such that f(p) = f(q), where a and b are some real numbers. Then, there exists at least one number c in (p, q) such that f'(c) = 0.

• Mean Value Theorem: Let f : [p, q] → R be a continuous function on [p, q] and differentiable on (p, q). Then, there exists at least one number c in (p, q) such that

$f^{\prime}(c)=\frac{f(p)-f(q)}{q-p}$

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