NCERT Solutions for Exercise 5.5 Class 12 Maths Chapter 5 - Continuity and Differentiability

Question:2. Differentiate the functions w.r.t. x.

$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Answer:

Given function is
$y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
Take log on both the sides

$\log y=\frac{1}{2}\log \left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right)$

$\log y=\frac{1}{2}(\log (x-1)+\log (x-2)-\log (x-3)-\log (x-4)-\log (x-5))$
Now, differentiation w.r.t. x is
$\frac{d(\log y)}{dx}=\frac{1}{2}(\frac{d(\log (x-1))}{dx}+\frac{d(\log (x-2))}{dx}-\frac{d(\log (x-3))}{dx}-\frac{d(\log (x-4))}{dx}-$$\frac{d(\log (x-5))}{dx})$

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

$\frac{dy}{dx}=y\cdot \frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

$\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

Therefore, the answer is $\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

Question:3 Differentiate the functions w.r.t. x. $(\log x{)}^{\cos x}$

Answer:

Given function is
$y=(\log x{)}^{\cos x}$
take log on both the sides
$\log y=\cos x\log (\log x)$
Now, differentiation w.r.t x is

$\frac{d(\log y)}{dx}=\frac{d(\cos x\log (\log x))}{dx}$

$\frac{1}{y}\cdot \frac{dy}{dx}=(-\sin x)(\log (\log x))+\cos x\cdot \frac{1}{\log x}\cdot \frac{1}{x}$

$\frac{dy}{dx}=y(\cos x\cdot \frac{1}{\log x}\cdot \frac{1}{x}-\sin x\log (\log x))$

$\frac{dy}{dx}=(\log x{)}^{\cos x}(\frac{\cos x}{x\log x}-\sin x\log (\log x))$

Therefore, the answer is $(\log x{)}^{\cos x}(\frac{\cos x}{x\log x}-\sin x\log (\log x))$

Question:4 Differentiate the functions w.r.t. x. $x^x-2^{\sin x}$

Answer:

Given function is
$y=x^x-2^{\sin x}$
Let's take $t=x^x$
take log on both the sides
$\log t=x\log x$
Now, differentiation w.r.t x is

$\log t=x\log x$

$\frac{d(\log t)}{dt}\cdot \frac{dt}{dx}=\frac{d(x\log x)}{dx}(\text{by chain rule})$

$\frac{1}{t}\cdot \frac{dt}{dx}=\log x+1$

$\frac{dt}{dx}=t(\log x+1)$

$\frac{dt}{dx}=x^x(\log x+1)(\because t=x^x)$

Similarly, take $k=2^{\sin x}$
Now, take log on both sides and differentiate w.r.t. x

$\log k=\sin x\log 2$

$\frac{d(\log k)}{dk}\cdot \frac{dk}{dx}=\frac{d(\sin x\log 2)}{dx}(\text{by chain rule})$

$\frac{1}{k}\cdot \frac{dk}{dx}=\cos x\log 2$

$\frac{dk}{dx}=k(\cos x\log 2)$

$\frac{dk}{dx}=2^{\sin x}(\cos x\log 2)(\because k=2^{\sin x})$

Now,
$$\begin{gathered} \frac{dy}{dx}=\frac{dt}{dx}-\frac{dk}{dx} \\ \frac{dy}{dx}=x^x(\log x+1)-2^{\sin x}(\cos x\log 2) \end{gathered}$$

Therefore, the answer is $x^x(\log x+1)-2^{\sin x}(\cos x\log 2)$

Question:5 Differentiate the functions w.r.t. x. $(x+3{)}^2.(x+4{)}^3.(x+5{)}^4$

Answer:

Given function is
$y=(x+3{)}^2.(x+4{)}^3.(x+5{)}^4$
Take log on both sides
$$\begin{gathered} \log y=\log [(x+3{)}^2.(x+4{)}^3.(x+5{)}^4] \\ \log y=2\log (x+3)+3\log (x+4)+4\log (x+5) \end{gathered}$$
Now, differentiate w.r.t. x we get,

$\frac{1}{y}\cdot \frac{dy}{dx}=2\cdot \frac{1}{x+3}+3\cdot \frac{1}{x+4}+4\cdot \frac{1}{x+5}$

$\frac{dy}{dx}=y(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5})$

$\frac{dy}{dx}=(x+3{)}^2(x+4{)}^3(x+5{)}^4(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5})$

$\frac{dy}{dx}=(x+3{)}^2(x+4{)}^3(x+5{)}^4\left(\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right)$

$\frac{dy}{dx}=(x+3)(x+4{)}^2(x+5{)}^3(9x^2+70x+133)$

Therefore, the answer is $(x+3)(x+4{)}^2(x+5{)}^3(9x^2+70x+133)$

Question:6 Differentiate the functions w.r.t. x. $(x+\frac{1}{x}{)}^x+x^{1+\frac{1}{x}}$

Answer:

Given function is
$y=(x+\frac{1}{x}{)}^x+x^{1+\frac{1}{x}}$
Let's take $t=(x+\frac{1}{x}{)}^x$
Now, take log on both sides
$\log t=x\log (x+\frac{1}{x})$
Now, differentiate w.r.t. x
we get,

$\frac{1}{t}\cdot \frac{dt}{dx}=\log (x+\frac{1}{x})+x(1-\frac{1}{x^2})\cdot \frac{1}{(x+\frac{1}{x})}$

$=\frac{x^2-1}{x^2+1}+\log (x+\frac{1}{x})$

$\frac{dt}{dx}=t(\frac{x^2-1}{x^2+1}+\log (x+\frac{1}{x}))$

$\frac{dt}{dx}={(x+\frac{1}{x})}^x(\frac{x^2-1}{x^2+1}+\log (x+\frac{1}{x}))$

Similarly, take $k=x^{1+\frac{1}{x}}$
Now, take log on both sides
$\log k=(1+\frac{1}{x})\log x$
Now, differentiate w.r.t. x
We get,

$\frac{1}{k}\cdot \frac{dk}{dx}=\frac{1}{x}(1+\frac{1}{x})+(-\frac{1}{x^2})\log x$

$=\frac{x^2+1}{x^2}+(-\frac{1}{x^2})\log x$

$\frac{dk}{dx}=k\left(\frac{x^2+1-\log x}{x^2}\right)$

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

Now,
$\frac{dy}{dx}=\frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}={(x+\frac{1}{x})}^x(\left(\frac{x^2-1}{x^2+1}\right)+\log (x+\frac{1}{x}))+x^{x+\frac{1}{x}}\left(\frac{x^2+1-\log x}{x^2}\right)$

Therefore, the answer is ${(x+\frac{1}{x})}^x(\left(\frac{x^2-1}{x^2+1}\right)+\log (x+\frac{1}{x}))+x^{x+\frac{1}{x}}\left(\frac{x^2+1-\log x}{x^2}\right)$

Question:7 Differentiate the functions w.r.t. x. $(\log x{)}^x+x^{\log x}$

Answer:

Given function is
$y=(\log x{)}^x+x^{\log x}$
Let's take $t=(\log x{)}^x$
Now, take log on both the sides
$\log t=x\log (\log x)$
Now, differentiate w.r.t. x
we get,

$\frac{1}{t}\cdot \frac{dt}{dx}=\log (\log x)+x\cdot \frac{1}{x}\cdot \frac{1}{\log x}=\log (\log x)+\frac{1}{\log x}$

$\frac{dt}{dx}=t\cdot (\log (\log x)+\frac{1}{\log x})$

$\frac{dt}{dx}=(\log x{)}^x\cdot \log (\log x)+(\log x{)}^x\cdot \frac{1}{\log x}$

$\frac{dt}{dx}=(\log x{)}^x\cdot \log (\log x)+(\log x{)}^{x-1}$

Similarly, take $k=x^{\log x}$
Now, take log on both sides
$\log k=\log x\log x=(\log x{)}^2$
Now, differentiate w.r.t. x
We get,
$$\begin{gathered} \frac{1}{k}\frac{dk}{dx}=2(\log x).\frac{1}{x} \\ \frac{dt}{dx}=k.(2(\log x).\frac{1}{x}) \\ \frac{dt}{dx}=x^{\log x}.(2(\log x).\frac{1}{x})=2x^{\log x-1}.\log x \end{gathered}$$
Now,
$\frac{dy}{dx}=\frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}=(\log x{)}^x(\log (\log x))+(\log x{)}^{x-1}+2x^{\log x-1}.\log x$
Therefore, the answer is $(\log x{)}^x(\log (\log x))+(\log x{)}^{x-1}+2x^{\log x-1}.\log x$

Question:8 Differentiate the functions w.r.t. x. $(\sin x{)}^x+{\sin }^{-1}\sqrt{x}$

Answer:

Given function is
$(\sin x{)}^x+{\sin }^{-1}\sqrt{x}$
Lets take $t=(\sin x{)}^x$
Now, take log on both the sides
$\log t=x\log (\sin x)$
Now, differentiate w.r.t. x
we get,

$\frac{1}{t}\cdot \frac{dt}{dx}=\log (\sin x)+x\cdot \cos x\cdot \frac{1}{\sin x}=\log (\sin x)+x\cdot \cot x(\because \frac{\cos x}{\sin x}=\cot x)$

$\frac{dt}{dx}=t\cdot (\log (\sin x)+x\cdot \cot x)$

$\frac{dt}{dx}=(\sin x{)}^x\cdot (\log (\sin x)+x\cdot \cot x)$

Similarly, take $k={\sin }^{-1}\sqrt{x}$
Now, differentiate w.r.t. x
We get,
$$\begin{gathered} \frac{dk}{dt}=\frac{1}{\sqrt{1-(\sqrt{x}{)}^2}}.\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x-x^2}} \\ \frac{dk}{dt}=\frac{1}{2\sqrt{x-x^2}} \end{gathered}$$
Now,
$\frac{dy}{dx}=\frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}=(\sin x{)}^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$
Therefore, the answer is $(\sin x{)}^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$

Question:9 Differentiate the functions w.r.t. x $x^{\sin x}+(\sin x{)}^{\cos x}$

Answer:

Given function is
$y=x^{\sin x}+(\sin x{)}^{\cos x}$
Now, take $t=x^{\sin x}$
Now, take log on both sides
$\log t=\sin x\log x$
Now, differentiate it w.r.t. x
we get,

$\frac{1}{t}\cdot \frac{dt}{dx}=\cos x\cdot \log x+\frac{1}{x}\cdot \sin x$

$\frac{dt}{dx}=t(\cos x\cdot \log x+\frac{1}{x}\cdot \sin x)$

$\frac{dt}{dx}=x^{\sin x}(\cos x\cdot \log x+\frac{1}{x}\cdot \sin x)$

Similarly, take $k=(\sin x{)}^{\cos x}$
Now, take log on both the sides
$\log k=\cos x\log (\sin x)$
Now, differentiate it w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx}=(-\sin x)(\log (\sin x))+\cos x\cdot \frac{1}{\sin x}\cdot \cos x=-\sin x\log (\sin x)+\cot x\cdot \cos x$
$\frac{dk}{dx}=k(-\sin x\log (\sin x)+\cot x\cdot \cos x)$
$\frac{dk}{dx}=(\sin x{)}^{\cos x}(-\sin x\log (\sin x)+\cot x\cdot \cos x)$
Now,
$\frac{dy}{dx}=x^{\sin x}(\cos x\log x+\frac{1}{x}.\sin x)+(\sin x{)}^{\cos x}(-\sin x\log (\sin x)+\cot x.\cos x)$
Therefore, the answer is $x^{\sin x}(\cos x\log x+\frac{1}{x}.\sin x)+(\sin x{)}^{\cos x}(-\sin x\log (\sin x)+\cot x.\cos x)$

Question:10 Differentiate the functions w.r.t. x. $x^{x\cos x}+\frac{x^2+1}{x^2-1}$

Answer:

Given function is
$x^{x\cos x}+\frac{x^2+1}{x^2-1}$
Take $t=x^{x\cos x}$
Take log on both the sides
$\log t=x\cos x\log x$
Now, differentiate w.r.t. x
we get,

$\frac{1}{t}\cdot \frac{dt}{dx}=\cos x\cdot \log x-x\cdot \sin x\cdot \log x+\frac{1}{x}\cdot x\cdot \cos x$

$\frac{dt}{dx}=t\cdot (\log x(\cos x-x\sin x)+\cos x)$

$\frac{dt}{dx}=x^{x\cos x}\cdot (\log x(\cos x-x\sin x)+\cos x)$

Similarly,
take $k=\frac{x^2+1}{x^2-1}$
Now. differentiate it w.r.t. x
we get,
$\frac{dk}{dx}=\frac{2x(x^2-1)-2x(x^2+1)}{(x^2-1{)}^2}=\frac{2x^3-2x-2x^3-2x}{(x^2-1{)}^2}=\frac{-4x}{(x^2-1{)}^2}$
Now,
$\frac{dy}{dx}=\frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}=x^{x\cos x}(\log x(\cos x-x\sin x)+\cos x)-\frac{4x}{(x^2-1{)}^2}$
Therefore, the answer is $x^{x\cos x}(\cos x(\log x+1)-x\sin x\log x)-\frac{4x}{(x^2-1{)}^2}$

Question:11 Differentiate the functions w.r.t. x. $(x\cos x{)}^x+(x\sin x{)}^{1/x}$

Answer:

Given function is
$f(x)=(x\cos x{)}^x+(x\sin x{)}^{1/x}$
Let's take $t=(x\cos x{)}^x$
Now, take log on both sides
$\log t=x\log (x\cos x)=x(\log x+\log \cos x)$
Now, differentiate w.r.t. x
we get,

$\frac{1}{t}\cdot \frac{dt}{dx}=(\log x+\log \cos x)+x(\frac{1}{x}+\frac{1}{\cos x}\cdot (-\sin x))$

$\frac{dt}{dx}=t(\log x+\log \cos x+1-x\tan x)(\because \frac{\sin x}{\cos x}=\tan x)$

$\frac{dt}{dx}=(x\cos x{)}^x(\log x+\log \cos x+1-x\tan x)$

$\frac{dt}{dx}=(x\cos x{)}^x(1-x\tan x+\log (x\cos x))$

Similarly, take $k=(x\sin x{)}^{\frac{1}{x}}$
Now, take log on both the sides
$\log k=\frac{1}{x}(\log x+\log \sin x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx}=\left(\frac{-1}{x^2}\right)(\log x+\log \sin x)+\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}\cdot \cos x)$
$\frac{dk}{dx}=\frac{k}{x^2}(-\log x-\log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})(\because \frac{\cos x}{\sin x}=\cot x)$
$\frac{dk}{dx}=\frac{(x\sin x{)}^{\frac{1}{x}}}{x^2}(-\log x-\log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})$
$\frac{dk}{dx}=(x\sin x{)}^{\frac{1}{x}}\cdot \frac{x\cot x+1-\log (x\sin x)}{x^2}$
Now,
$\frac{dy}{dx}=\frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}=(x\cos x{)}^x(+1-x\tan x+\log (x\cos x))+(x\sin x{)}^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Therefore, the answer is $(x\cos x{)}^x(1-x\tan x+\log (x\cos x))+(x\sin x{)}^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$

Question:12 Find dy/dx of the functions given in Exercises 12 to 15

$x^y+y^x=1$.

Answer:

Given function is
$f(x)=x^y+y^x=1$
Now, take $t=x^y$
take log on both sides
$\log t=y\log x$
Now, differentiate w.r.t x
we get,

$\frac{1}{t}\cdot \frac{dt}{dx}=\frac{dy}{dx}\cdot (\log x)+y\cdot \frac{1}{x}=\frac{dy}{dx}\cdot (\log x)+\frac{y}{x}$

$\frac{dt}{dx}=t(\frac{dy}{dx}\cdot (\log x)+\frac{y}{x})$

$\frac{dt}{dx}=x^y(\frac{dy}{dx}\cdot (\log x)+\frac{y}{x})$

Similarly, take $k=y^x$
Now, take log on both sides
$\log k=x\log y$
Now, differentiate w.r.t. x
we get,
$$\begin{gathered} \frac{1}{k}\frac{dk}{dx}=(\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx} \\ \frac{dk}{dx}=k(\log y+\frac{x}{y}\frac{dy}{dx}) \\ \frac{dk}{dx}=(y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) \end{gathered}$$
Now,
$f^{{}^{\prime }}(x)=\frac{dt}{dx}+\frac{dk}{dx}=0$

$(x^y)(\frac{dy}{dx}\log x+\frac{y}{x})+(y^x)(\log y+\frac{x}{y}\frac{dy}{dx})=0$

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

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

Therefore, the answer is $\frac{-(y{x}^{y-1}+y^x(\log y))}{(x^y(\log x)+x{y}^{x-1})}$

Question:13 Find dy/dx of the functions given in Exercises 12 to 15.

$y^x=x^y$

Answer:

Given function is
$f(x)\Rightarrow x^y=y^x$
Now, take $t=x^y$
take log on both sides
$\log t=y\log x$
Now, differentiate w.r.t x
we get,

$\frac{1}{t}\cdot \frac{dt}{dx}=\frac{dy}{dx}\cdot \log x+y\cdot \frac{1}{x}=\frac{dy}{dx}\cdot \log x+\frac{y}{x}$

$\frac{dt}{dx}=t(\frac{dy}{dx}\cdot \log x+\frac{y}{x})$

$\frac{dt}{dx}=x^y(\frac{dy}{dx}\cdot \log x+\frac{y}{x})$

Similarly, take $k=y^x$
Now, take log on both sides
$\log k=x\log y$
Now, differentiate w.r.t. x
we get,
$$\begin{gathered} \frac{1}{k}\frac{dk}{dx}=(\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx} \\ \frac{dk}{dx}=k(\log y+\frac{x}{y}\frac{dy}{dx}) \\ \frac{dk}{dx}=(y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) \end{gathered}$$
Now,
$f^{{}^{\prime }}(x)\Rightarrow \frac{dt}{dx}=\frac{dk}{dx}$

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

Therefore, the answer is $\frac{x}{y}\left(\frac{y-x\log y}{x-y\log x}\right)$

Question:14 Find dy/dx of the functions given in Exercises 12 to 15. $(\cos x{)}^y=(\cos y{)}^x$

Answer:

Given function is
$f(x)\Rightarrow (\cos x{)}^y=(\cos y{)}^x$
Now, take log on both the sides
$y\log \cos x=x\log \cos y$
Now, differentiate w.r.t x
$\frac{dy}{dx}(\log \cos x)-y\tan x=\log \cos y-x\tan y\frac{dy}{dx}$
By taking similar terms on the same side
We get,
$(\frac{dy}{dx}(\log \cos x)-y\tan x)=(\log \cos y-x\tan y\frac{dy}{dx})$
$\frac{dy}{dx}(\log \cos x+x\tan y)=\log \cos y+y\tan x$
$\frac{dy}{dx}=\frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

Therefore, the answer is $\frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

Question:15 Find dy/dx of the functions given in Exercises 12 to 15. $xy=e^{x-y}$

Answer:

Given function is
$f(x)\Rightarrow xy=e^{x-y}$
Now, take log on both the sides

$\log x+\log y=(x-y)(1)(\because \log e=1)$

$\log x+\log y=x-y$

Now, differentiate w.r.t x
$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$
By taking similar terms on same side
We get,
$$\begin{gathered} (\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x} \\ \frac{y+1}{y}.\frac{dy}{dx}=\frac{x-1}{x} \\ \frac{dy}{dx}=\frac{y}{x}.\frac{x-1}{y+1} \end{gathered}$$
Therefore, the answer is $\frac{y}{x}.\frac{x-1}{y+1}$