Chapter 5: Continuity and Differentiability

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} $

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} $

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} $

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)} $

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} $

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)} $

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} $

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} $

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}}} $

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. $x$.

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

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

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

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