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

Question:2. Find dy/dx in the following: $2x+3y=\sin y$

Answer:

Given function is
$2x+3y=\sin y$
We can rewrite it as
$\sin y-3y=2x$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(\sin y-3y)=\frac{d(2x)}{dx}$

$(\cos y\frac{dy}{dx}-3\frac{dy}{dx})=2$
$\frac{dy}{dx}=\frac{2}{\cos y-3}$
Therefore, the answer is $\frac{2}{\cos y-3}$

Question:3. Find dy/dx in the following: $ax+b{y}^2=\cos y$

Answer:

Given function is
$ax+b{y}^2=\cos y$
We can rewrite it as
$b{y}^2-\cos y=-ax$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(2by-(-\sin y))=\frac{d(-ax)}{dx}=-a$
$\frac{dy}{dx}=\frac{-a}{2by+\sin y}$
Therefore, the answer is $\frac{-a}{2by+\sin y}$

Question:4. Find dy/dx in the following:

$xy+y^2=\tan x+y$

Answer:

Given function is
$xy+y^2=\tan x+y$
We can rewrite it as
$xy+y^2-y=\tan x$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y-1)=\frac{d(\tan x)}{dx}={\sec }^{2}x$
$\frac{dy}{dx}=\frac{{\sec }^{2}x-y}{x+2y-1}$
Therefore, the answer is $\frac{{\sec }^{2}x-y}{x+2y-1}$

Question:5. Find dy/dx in the following: $x^2+xy+y^2=100$

Answer:

Given function is
$x^2+xy+y^2=100$
We can rewrite it as
$xy+y^2=100-x^2$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y)=\frac{d(100-x^2)}{dx}=-2x$
$\frac{dy}{dx}=\frac{-2x-y}{x+2y}$
Therefore, the answer is $\frac{-2x-y}{x+2y}$

Question:6 Find dy/dx in the following:

$x^3+x^{2}y+x{y}^2+y^3=81$

Answer:

Given function is
$x^3+x^{2}y+x{y}^2+y^3=81$
We can rewrite it as
$x^{2}y+x{y}^2+y^3=81-x^3$
Now, differentiation w.r.t. x is
$\frac{d(x^{2}y+x{y}^2+y^3)}{dx}=\frac{d(81-x^3)}{dx}$
$$\begin{gathered} 2xy+y^2+\frac{dy}{dx}(x^2+2xy+3y^2)=-3x^2 \\ \frac{dy}{dx}=\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2} \end{gathered}$$
Therefore, the answer is $\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$

Question:7. Find dy/dx in the following: ${\sin }^{2}y+\cos xy=k$

Answer:

Given function is
${\sin }^{2}y+\cos xy=k$
Now, differentiation w.r.t. x is
$\frac{d({\sin }^{2}y+\cos xy)}{dx}=\frac{d(k)}{dx}$
$$\begin{gathered} 2\sin y\cos y\frac{dy}{dx}+(-\sin xy)(y+x\frac{dy}{dx})=0 \\ \frac{dy}{dx}(2\sin y\cos y-x\sin xy)=y\sin xy \\ \frac{dy}{dx}=\frac{y\sin xy}{2\sin y\cos y-x\sin xy}=\frac{y\sin xy}{\sin 2y-x\sin xy}(\because 2\sin x\cos y=\sin 2x) \end{gathered}$$
Therefore, the answer is $\frac{y\sin xy}{\sin 2y-x\sin xy}$

Question:8. Find dy/dx in the following:

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

Answer:

Given function is
${\sin }^{2}x+{\cos }^{2}y=1$
We can rewrite it as
${\cos }^{2}y=1-{\sin }^{2}x$
Now, differentiation w.r.t. x is
$\frac{d({\cos }^{2}y)}{dx}=\frac{d(1-{\sin }^{2}x)}{dx}$
$$\begin{gathered} 2\cos y(-\sin y)\frac{dy}{dx}=-2\sin x\cos x \\ \frac{dy}{dx}=\frac{2\sin x\cos x}{2\sin y\cos y}=\frac{\sin 2x}{\sin 2y}(\because 2\sin a\cos a=\sin 2a) \end{gathered}$$
Therefore, the answer is $\frac{\sin 2x}{\sin 2y}$

Question:9 Find dy/dx in the following:

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

Answer:

Given function is
$y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
Lets consider $x=\tan t$
Then,
$\frac{d(x)}{dx}=\frac{d(\tan t)}{dt}.\frac{dt}{dx}(bychainrule)$
$$\begin{gathered} 1={\sec }^{2}t.\frac{dt}{dx} \\ \frac{dt}{dx}=\frac{1}{{\sec }^{2}t}=\frac{1}{1+{\tan }^{2}t}=\frac{1}{1+x^2}(\because {\sec }^{2}t=1+{\tan }^{2}tandx=\tan t) \end{gathered}$$
Now,
$\frac{2x}{1+x^2}=\frac{2\tan t}{1+{\tan }^{2}t}=\sin 2t(\because \sin 2x=\frac{2\tan x}{1+{\tan }^{2}x})$
Our equation reduces to
$y={\sin }^{-1}(\sin 2t)$
$y=2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx}=\frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx}=2.\frac{1}{1+x^2}=\frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$

Question:10. Find dy/dx in the following:
$y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Answer:

Given function is
$y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$
Lets consider $x=\tan t$
Then,
$\frac{d(x)}{dx}=\frac{d(\tan t)}{dt}.\frac{dt}{dx}(bychainrule)$
$$\begin{gathered} 1={\sec }^{2}t.\frac{dt}{dx} \\ \frac{dt}{dx}=\frac{1}{{\sec }^{2}t}=\frac{1}{1+{\tan }^{2}t}=\frac{1}{1+x^2}(\because {\sec }^{2}t=1+{\tan }^{2}tandx=\tan t) \end{gathered}$$
Now,
$\frac{3x-x^3}{1-3x^2}=\frac{3\tan t-{\tan }^{3}t}{1-3{\tan }^{2}t}=\tan 3t(\because \tan 3x=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x})$
Our equation reduces to
$y={\tan }^{-1}(\tan 3t)$
$y=3t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx}=\frac{d(3t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx}=3.\frac{1}{1+x^2}=\frac{3}{1+x^2}$
Therefore, the answer is $\frac{3}{1+x^2}$

Question:11. Find dy/dx in the following:

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

Answer:

Given function is
$y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
Let's consider $x=\tan t$
Then,
$\frac{d(x)}{dx}=\frac{d(\tan t)}{dt}.\frac{dt}{dx}(bychainrule)$
$$\begin{gathered} 1={\sec }^{2}t.\frac{dt}{dx} \\ \frac{dt}{dx}=\frac{1}{{\sec }^{2}t}=\frac{1}{1+{\tan }^{2}t}=\frac{1}{1+x^2}(\because {\sec }^{2}t=1+{\tan }^{2}tandx=\tan t) \end{gathered}$$
Now,
$\frac{1-x^2}{1+x^2}=\frac{1-{\tan }^{2}t}{1+{\tan }^{2}t}=\cos 2t(\because \cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x})$
Our equation reduces to
$y={\cos }^{-1}(\cos 2t)$
$y=2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx}=\frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx}=2.\frac{1}{1+x^2}=\frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$

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

Answer:

Given function is
$y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
We can rewrite it as
$\sin y=\left(\frac{1-x^2}{1+x^2}\right)$
Let's consider $x=\tan t$
Then,
$\frac{d(x)}{dx}=\frac{d(\tan t)}{dt}.\frac{dt}{dx}(bychainrule)$
$$\begin{gathered} 1={\sec }^{2}t.\frac{dt}{dx} \\ \frac{dt}{dx}=\frac{1}{{\sec }^{2}t}=\frac{1}{1+{\tan }^{2}t}=\frac{1}{1+x^2}(\because {\sec }^{2}t=1+{\tan }^{2}tandx=\tan t) \end{gathered}$$
Now,
$\frac{1-x^2}{1+x^2}=\frac{1-{\tan }^{2}t}{1+{\tan }^{2}t}=\cos 2t(\because \cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x})$
Our equation reduces to
$\sin y=\cos 2t$
Now, differentiation w.r.t. x is
$\frac{d(\sin y)}{dx}=\frac{d(\cos 2t)}{dt}.\frac{dt}{dx}$
$\cos y\frac{dy}{dx}=2(-\sin 2t).\frac{1}{1+x^2}=\frac{-2\sin 2t}{1+x^2}$$=\frac{-2.\frac{2\tan t}{1+{\tan }^{2}t}}{1+x^2}=\frac{-2.\frac{2x}{1+x^2}}{1+x^2}=\frac{-4x}{(1+x^2{)}^2}$
$(\because \sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}andx=\tan t)$
$\sin y=\left(\frac{1-x^2}{1+x^2}\right)\Rightarrow \cos y=\frac{2x}{1+x^2}$
$\frac{2x}{1+x^2}\frac{dy}{dx}=\frac{-4x}{(1+x^2{)}^2}$
$\frac{dy}{dx}=\frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$

Question:13. Find dy/dx in the following:

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

Answer:

Given function is
$y={\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$
We can rewrite it as
$\cos y=\left(\frac{2x}{1+x^2}\right)$
Let's consider $x=\tan t$
Then,
$\frac{d(x)}{dx}=\frac{d(\tan t)}{dt}.\frac{dt}{dx}(bychainrule)$
$$\begin{gathered} 1={\sec }^{2}t.\frac{dt}{dx} \\ \frac{dt}{dx}=\frac{1}{{\sec }^{2}t}=\frac{1}{1+{\tan }^{2}t}=\frac{1}{1+x^2}(\because {\sec }^{2}t=1+{\tan }^{2}tandx=\tan t) \end{gathered}$$
Now,
$\frac{2x}{1+x^2}=\frac{2\tan t}{1+{\tan }^{2}t}=\sin 2t(\because \sin 2x=\frac{2\tan x}{1+{\tan }^{2}x})$
Our equation reduces to
$\cos y=\sin 2t$
Now, differentiation w.r.t. x is
$\frac{d(\cos y)}{dx}=\frac{d(\sin 2t)}{dt}.\frac{dt}{dx}$
$(-\sin y)\frac{dy}{dx}=2(\cos 2t).\frac{1}{1+x^2}=\frac{2\cos 2t}{1+x^2}$$=\frac{2.\frac{1-{\tan }^{2}t}{1+{\tan }^{2}t}}{1+x^2}=\frac{2.\frac{1-x^2}{1+x^2}}{1+x^2}=\frac{2(1-x^2)}{(1+x^2{)}^2}$
$(\because \cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}andx=\tan t)$
$\cos y=\left(\frac{2x}{1+x^2}\right)\Rightarrow \sin y=\frac{1-x^2}{1+x^2}$
$-\frac{1-x^2}{1+x^2}\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2{)}^2}$
$\frac{dy}{dx}=\frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$

Question:14. Find dy/dx in the following:

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

Answer:

Given function is
$y={\sin }^{-1}(2x\sqrt{1-x^2})$
Lets take $x=\sin t$
Then,
$\frac{d(x)}{dx}=\frac{(\sin t)}{dt}.\frac{dt}{dx}(bychainrule)$
$1=\cos t.\frac{dt}{dx}$
$\frac{dt}{dx}=\frac{1}{\cos t}=\frac{1}{\sqrt{1-{\sin }^{2}t}}=\frac{1}{\sqrt{1-x^2}}$
$(\because \cos x=\sqrt{1-{\sin }^{2}x}andx=\sin t)$
And
$2x\sqrt{1-x^2}=2\sin t\sqrt{1-{\sin }^{2}t}=2\sin t\sqrt{{\cos }^{2}t}=2\sin t\cos t=\sin 2t$
$(\because \cos x=\sqrt{1-{\sin }^{2}x}and2\sin x\cos x=\sin 2x)$
Now, our equation reduces to
$y={\sin }^{-1}(\sin 2t)$
$y=2t$
Now, differentiation w.r.t. x
$\frac{d(y)}{dx}=\frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx}=2.\frac{1}{\sqrt{1-x^2}}=\frac{2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{2}{\sqrt{1-x^2}}$

Question:15. Find dy/dx in the following:

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

Answer:

Given function is
$y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$
Let's take $x=\cos t$
Then,
$\frac{d(x)}{dx}=\frac{(\cos t)}{dt}.\frac{dt}{dx}(bychainrule)$
$1=-\sin t.\frac{dt}{dx}$
$\frac{dt}{dx}=\frac{-1}{\sin t}=\frac{-1}{\sqrt{1-{\cos }^{2}t}}=\frac{-1}{\sqrt{1-x^2}}$
$(\because \sin x=\sqrt{1-{\cos }^{2}x}andx=\cos t)$
And
$\frac{1}{2x^2-1}=\frac{1}{2{\cos }^{2}t-1}=\frac{1}{\cos 2t}=\sec 2t$
$(\because \cos 2x=\sqrt{2{\cos }^{2}x-1}and\cos x=\frac{1}{\sec x})$

Now, our equation reduces to
$y={\sec }^{-1}(\sec 2t)$
$y=2t$
Now, differentiation w.r.t. x
$\frac{d(y)}{dx}=\frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx}=2.\frac{-1}{\sqrt{1-x^2}}=\frac{-2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{-2}{\sqrt{1-x^2}}$