NCERT Solutions for Class 12 Maths Chapter 5 Continuity And Differentiability Ex 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 $ 

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

Differentiating the above relationship with respect to

x, 

We get

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}}}} $ 

Conclusion

NCERT Solutions for Class 12 Maths Ex 5.3 Continuity and Differentiability by Vedantu are essential for mastering the concepts of differentiability. This exercise emphasizes understanding and applying various differentiation techniques such as the product rule, quotient rule, chain rule, implicit differentiation, and higher-order derivatives. These skills are crucial for solving complex problems in board exams and competitive exams like JEE. In previous year's question papers, there were around 3-4 questions based on these topics, highlighting their importance in exams. Paying close attention to the detailed solutions provided by Vedantu will help you gain confidence and proficiency in handling a wide range of differentiation problems.

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