If $$y^x=x^y$$, then find $$\displaystyle\frac{dy}{dx}$$.

$$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$

$$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$$

$$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$$

If $$\displaystyle y^{x}=x^{\sin y} $$, find $$\cfrac{dy}{dx}$$.

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]$$

$$\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]$$

If $$y = \displaystyle (\tan x)^{\log x}$$, then $$\cfrac{dy}{dx} = $$

$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx$$

$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } $$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 2

Find

$\displaystyle\frac{dy}{dx}$ if

$x=a(\theta-\sin{\theta})$ and

$y=a(1-\cos{\theta})$ .

0%
$\cot{\left(\displaystyle\frac{\theta}{2}\right)}$
0%
$\tan{\left(\displaystyle\frac{\theta}{2}\right)}$
0%
$-\cos{\left(\displaystyle\frac{\theta}{2}\right)}$
0%
None of these

Explanation

We have

$x=a(\theta-\sin{\theta})$ and

$y=a(1-\cos{\theta})$
Differentiate w.r.t

$\theta$

$\therefore\displaystyle\frac{dx}{d\theta}=a(1-\cos{\theta})$ and

$\displaystyle\frac{dy}{d\theta}=a\sin{\theta}$
or

$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{d\theta}}{\displaystyle\frac{dx}{d\theta}}$

$=\displaystyle\frac{a\sin{\theta}}{a(1-\cos{\theta})}=\frac{2\sin{\left(\displaystyle\frac{\theta}{2}\right)}\cos{\left(\displaystyle\frac{\theta}{2}\right)}}{2\sin^{2}{\left(\displaystyle\frac{\theta}{2}\right)}}=\cot{\left(\displaystyle\frac{\theta}{2}\right)}$

The derivative of

$f(\tan x)$ w.r.t.

$g (\sec x)$ at

$x=\displaystyle \frac{\pi }{4},$ where

${f}'(1)=2$ and

${g}'(\sqrt{2})=4,$ is

0%
$\displaystyle \frac{1}{\sqrt{2}}$
0%
$\sqrt{2}$
0%
$1$
0%
none of these

Explanation

$y = f(\tan x)$ and

$z = g (\sec x)$

$\dfrac{dy}{dx}= f'(\tan x) . \sec^{2}x$ and

$\dfrac{dz}{dx}= g'(\sec x) . \sec x \tan x$

$\therefore \dfrac{dy}{dz}= \dfrac{f'(\tan x).\sec^{2} x}{g'(\sec x).\sec x \tan x}$

$\left.\begin{matrix}\dfrac{dy}{dz}\end{matrix}\right|_{x=\tfrac{\pi }{4}}=\dfrac{f'(1).(\sqrt{2})^{2}}{g'(\sqrt{2}).(\sqrt{2})}=\dfrac{1}{\sqrt{2}}$

$r=\sqrt{x^{2}+y^{2}+z^{2}}$ and

$x=2\sin 3t,y=2\cos 3t,z=8t$ then

$\displaystyle \frac{dr}{dt}=$

0%
$\displaystyle \frac{32t}{\sqrt{1+16t^{2}}}$
0%
$\displaystyle \frac{16t}{\sqrt{1+16t^{2}}}$
0%
$\displaystyle \frac{t}{\sqrt{1+16t^{2}}}$
0%
$\displaystyle \frac{4t}{\sqrt{1+16t^{2}}}$

Explanation

Given :

$x=2\sin 3t,y=2\cos 3t,z=8t$

$r=\sqrt{x^{2}+y^{2}+z^{2}}$

$=\sqrt{(2\sin 3t)^{2}+(2 \cos 3t)^{2}+(8t)^{2}}$

$=\sqrt{4\sin^{2}3t + 4\cos^{2}3t+64t^{2}}$

$=\sqrt{4+64t^{2}}$

$\implies r^{2}=4+64t^{2}$

Differentiating above equation we get

$2r\dfrac{dr}{dt}=2\times 64\times t$

$\implies \dfrac{dr}{dt}=\dfrac{64\times t}{r}$

$\implies \dfrac{dr}{dt}=\dfrac{64\times t}{\sqrt{4+64t^{2}}}=\dfrac{64t}{2\sqrt{1+16t^{2}}}=\dfrac{32t}{\sqrt{1+16t^{2}}}$

$y^x=x^y$ , then find

$\displaystyle\frac{dy}{dx}$ .

0%
$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$
0%
$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$
0%
${y(x\log{y}-y)}$
0%
$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$

Explanation

$y^x=x^y$

Taking log on both sides

$\Rightarrow$

$\log{y^x}=\log{x^y}$

$\Rightarrow$

$x\log{y}=y\log{x}$

$\Rightarrow$

$\displaystyle x\frac{1}{y}\frac{dy}{dx}+\log{y}\times 1=\frac{y}{x}+\log{x}\frac{dy}{dx}$

$\Rightarrow$

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

$\Rightarrow$

$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{y}{x}-\log{y}}{\displaystyle\frac{x}{y}-\log{x}}=\frac{y(y-x\log{y})}{x(x-y\log{x})}=\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$

$\displaystyle x=\frac{2t}{1+t^2}$ ,

$\displaystyle y=\frac{1-t^2}{1+t^2}$ , then

$\displaystyle\frac{dy}{dx}$ at

$t=2$ is

0%
$\displaystyle\frac{4}{3}$
0%
$\displaystyle -\frac{4}{3}$
0%
$\displaystyle\frac{3}{4}$
0%
$\displaystyle -\frac{3}{4}$

Explanation

Given that

$\displaystyle x=\frac{2t}{1+t^2}$ ,

$\displaystyle y=\frac{1-t^2}{1+t^2}$
By differentiating w.r. to

$t$ , we get

$\displaystyle \frac { dx }{ dt } =\frac { \left( \left( 1+{ t }^{ 2 } \right)\displaystyle \frac { d }{ dt } \left( 2t \right) -2t\displaystyle \frac { d }{ dt } \left( 1+{ t }^{ 2 } \right) \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } }$
And

$\displaystyle \frac { dy }{ dt } =\frac { \left( \left( 1+{ t }^{ 2 } \right) \displaystyle \frac { d }{ dt } \left( 1-{ t }^{ 2 } \right) -\left( 1-{ t }^{ 2 } \right) \displaystyle \frac { d }{ dt } \left( 1+{ t }^{ 2 } \right) \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } }$

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

$\displaystyle\frac{dy}{dt}=\frac{(1+t^2)(-2t)-(1-t^2)2t}{{(1+t^2)}^2}=\frac{-4t}{{(1+t^2)}^2}$

$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{-4t}{2-2t^2}=\frac{2t}{t^2-1}$

$\therefore { \left( \displaystyle \frac { dy }{ dx } \right) }_{ t=2 }=\displaystyle \frac { 4 }{ 3 }$

$\displaystyle\frac{dy}{dx}$ for

$y=x^x$ is

0%
$x^x(1-\log{x})$
0%
$x^x(1-\log{y})$
0%
$x^x(1+\log{y})$
0%
$x^x(1+\log{x})$

Explanation

$y=x^x$
Taking

$\log$ on both the sides

$\log { y } =\log { \left( { x }^{ x } \right) }$

$\log { y } =x\log { x }$
Differentiate w.r to

$x$

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

$\displaystyle \frac { dy }{ dx } =y\left[ \frac { x }{ x } +\log { x } \right]$

$\displaystyle \frac { dy }{ dx } ={ x }^{ x }\left[ 1+\log { x } \right]$

Let the function

$y=f(x)$ be given by

$x=t^{5}-5t^{3}-20t+7$ and

$y=4t^{3}-3t^{2}-18t+3$ , where

$t\epsilon \left ( -2, 2 \right )$ . Then

$f^{'}(x)$ at

$t=1$ is ?

0%
$\displaystyle \frac{5}{2}$
0%
$\displaystyle \frac{2}{5}$
0%
$\displaystyle \frac{7}{5}$
0%
none of these

Explanation

Given,

$x=t^{5}-5t^{3}-20t+7$

$\displaystyle \frac{dx}{dt}=5t^{4}-15t^{2}-20$

$\displaystyle \left (\frac{dx}{dt}\right)_{t=1}=-30$
Also, given

$y=4t^{3}-3t^{2}-18t+3$

$\displaystyle \frac{dy}{dt}=12t^{2}-6t-18$

$\displaystyle \left (\frac{dy}{dt}\right)_{t=1}=-12$
So,

$\displaystyle \left (\frac{dy}{dx}\right)_{t=1}=\frac{2}{5}$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

$\displaystyle f\left ( x \right )= g\left ( x \right ).\sin x.$

$f'\left( x \right)=g\left( x \right)\cos { x } +g'\left( x \right)\sin { x }$

Putting

$x=0$

$f'\left( 0 \right)=g\left( 0 \right)=0$

Now,

$\displaystyle f'\left( 0 \right)=\lim _{ x\rightarrow 0 }{ \dfrac { f'\left( x \right)-f'\left( 0 \right) }{ x } }$

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } +g'\left( x \right)\sin { x } -g\left( 0 \right) }{ x } }$

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ x } } +\lim _{ x\rightarrow 0 }{ \dfrac { g'\left( x \right)\sin { x } }{ x } }$

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ x } }$

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ \sin { x } } }$

$=\lim _{ x\rightarrow 0 }{ \left( g\left( x \right)\cot { x-g\left( 0 \right)\csc { x } } \right) }$

$\displaystyle\frac{dy}{dx}$ at

$\displaystyle t=\frac{\pi}{4}$ for

$\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]$ and

$y=a\sin{t}$ is

0%
$1$
0%
$0$
0%
$-1$
0%
$\displaystyle \frac { 1 }{ 2 }$

Explanation

$\displaystyle x=\frac{\pi}{4}$ for

$\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]$ and

$y=a\sin{t}$

Differentiating w.r.t

$t$ , we get

$\displaystyle \frac { dx }{ dt } =a\left[ \frac { d }{ dt } \cos { t } +\frac { 1 }{ 2 } \frac { d }{ dt } \left( \log { \tan ^{ 2 }{ \frac { t }{ 2 } } } \right) \right]$

$\displaystyle\frac{dx}{dt}=a\left[-\sin{t}+\frac{1}{\displaystyle\tan{\frac{t}{2}}}\sec^2{\frac{t}{2}}\times\frac{1}{2}\right]$

$\displaystyle \frac { dx }{ dt } =a\left[ -\sin { t } +\frac { 1 }{ 2\displaystyle \frac { \sin { { t }/{ 2 } } }{ \cos { { t }/{ 2 } } } .\cos ^{ 2 }{ { t }/{ 2 } } } \right]$

$\displaystyle =a\left[-\sin{t}+\frac{1}{\displaystyle 2\sin{\frac{t}{2}}\cos{\frac{t}{2}}}\right]$

$\displaystyle =a\left[-\sin{t}+\frac{1}{\sin{t}}\right]$

$=a\left[ \displaystyle \frac { 1-\sin ^{ 2 }{ t } }{ \sin { t } } \right] =a\left[ \displaystyle \frac { \cos ^{ 2 }{ t } }{ \sin { t } } \right]$

$\displaystyle\frac{dy}{dt}=a\cos{t}$

$\displaystyle\therefore\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{a\cos{t}}{\displaystyle\frac{a\cos^2{t}}{\sin{t}}}=\tan{t}$
At

$\displaystyle t=\frac{\pi}{4}$ ,

$\displaystyle\frac{dy}{dx}=1$

The relation between the parameter '

$t$ ' and the angle

$\alpha$ between the tangent to the given curve and the

$x-$ axis is given by, '

$t$ ' equals to

0%
$\displaystyle \dfrac {\pi}{2}-\alpha$
0%
$\displaystyle \dfrac {\pi}{4}+\alpha$
0%
$\alpha-\displaystyle \dfrac {\pi}{4}$
0%
$\displaystyle \dfrac {\pi}{4}-\alpha$

Explanation

$x=e^t \cos t$ and

$y=e^t \sin t$

$\displaystyle \dfrac { dy }{ dt } =\displaystyle \dfrac { d }{ dt } \left( { e }^{ t }\sin { t } \right)$

$\displaystyle \dfrac { dy }{ dt } ={ e }^{ t }\left( \cos { t } +\sin { t } \right)$

$\displaystyle \dfrac { dx }{ dt } =\displaystyle \dfrac { d }{ dt } \left( { e }^{ t }\cos { t } \right)$

$\displaystyle \dfrac { dx }{ dt } ={ e }^{ t }\left( \cos { t } -\sin { t } \right)$

$\therefore \displaystyle \dfrac { dy }{ dx } =\displaystyle \dfrac { \cos { t } +\sin { t } }{ \cos { t } -\sin { t } }$

$\therefore \tan { \left( \displaystyle \dfrac { \pi }{ 4 } +t \right) =\tan { \alpha } }$

$\displaystyle \dfrac { \pi }{ 4 } +t=\alpha$

$t=\alpha -\displaystyle \dfrac { \pi }{ 4 }$

$y={(\tan{x})}^{\displaystyle{(\tan{x})}^{\displaystyle\tan{x}}}$ , then find

$\displaystyle\frac{dy}{dx}$ at

$\displaystyle x=\frac{\pi}{4}$ .

0%
$0$
0%
$1$
0%
$-1$
0%
$2$

Explanation

Taking

$\log$ on both sides, we get

$\log{y}={(\tan{x})}^{\displaystyle\tan{x}}\log{\tan{x}}$
Again taking

$\log$ on both the sides

$\log{\log{y}}=[\tan{x}\log{\tan{x}}]+\log{\log{\tan{x}}}$
Differentiating w.r.t.

$x$ , we get

$\displaystyle\frac{1}{y\log{y}}\frac{dy}{dx}=\log { \tan { x } \frac { d }{ dx } \left( \tan { x } \right) +\tan { x } \frac { d }{ dx } \left( \log { \tan { x } } \right) +\frac { d }{ dx } \left( \log { \log { \tan { x } } } \right) }$

$\displaystyle \frac{dy}{dx}=\log { \tan { x } .\sec ^{ 2 }{ x } +\tan { x } .\frac { \sec ^{ 2 }{ x } }{ \tan { x } } +\frac { \sec ^{ 2 }{ x } }{ \tan { x } .\log { \tan { x } } } }$

$\displaystyle \frac { dy }{ dx } =y\log { y.\sec ^{ 2 }{ x } \left[ \log { \tan { x } +1+\frac { 1 }{ \tan { x } .\log { \tan { x } } } } \right] }$
At

$\displaystyle x=\frac{\pi}{4}$ ,

$y=1$ and

$\log{y}=0$
So, putting this values in the equation of

$\displaystyle \frac { dy }{ dx }$ , we get

$\displaystyle\therefore{\left(\frac{dy}{dx}\right)}_{\displaystyle x=\frac{\pi}{4}}=0$

$x=a\sec^{3}{\theta}$ and

$y=a\tan^{3}{\theta}$ , then

$\displaystyle\frac{dy}{dx}$ at

$\theta=\displaystyle\frac{\pi}{3}$ is

0%
$\displaystyle\frac{\sqrt{3}}{2}$
0%
$\displaystyle\frac{\sqrt{5}}{2}$
0%
$\displaystyle\frac{\sqrt{2}}{3}$
0%
$\displaystyle\frac{\sqrt{4}}{3}$

Explanation

We have

$x=a\sec^{3}{\theta}$ and

$y=a\tan^{3}{\theta}$
Differentiate w. r to

$\theta$

$\displaystyle\frac{dx}{d\theta}=3a\sec^{2}{\theta}\displaystyle\frac{d}{d\theta}(\sec{\theta})=3a\sec^{3}{\theta}\tan{\theta}$

$\displaystyle\frac{dy}{d\theta}=3a\tan^{2}{\theta}\displaystyle\frac{d}{d\theta}(\tan{\theta})=3a\tan^{2}{\theta}\sec^{2}{\theta}$

$\therefore \displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{d\theta}}{\displaystyle\frac{dx}{d\theta}}=\frac{3a\tan^{2}{\theta}\sec^{2}{\theta}}{3a\sec^{3}{\theta}\tan{\theta}}=\frac{\tan{\theta}}{\sec{\theta}}=\sin{\theta}$

$\displaystyle{\left(\frac{dy}{dx}\right)}_{\displaystyle\theta=\frac{\pi}{3}}=\sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2}$

$\displaystyle y^{x}=x^{\sin y}$ , find

$\cfrac{dy}{dx}$ .

0%
$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$
0%
$\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]$
0%
$\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$
0%
$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]$

Explanation

Given

$\displaystyle y^{x}\:=x^{\sin \:y}$ .

Take log both sides.

$\displaystyle x \log y = \sin y \: \log \:x.$

Differentiate w.r.t. x

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

$\displaystyle \left ( \log y - \frac{\sin y}{x} \right )=\frac{dy}{dx}\left [ \cos y \log x-\frac{x}{y} \right ]$

$\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}\left [ \frac{x \log y-\sin y}{y \log x \cos y-x}\right ]$

Discuss the applicability of Rolle's theorem to

$\displaystyle f(x)=\log \left[\frac{x^{2}+ab}{(a+b)x}\right],$ in the interval

$[a,b].$

0%
Yes Rolle's theorem is applicable and the stationary point is $x=\sqrt { ab }$
0%
No Rolle's theorem is not applicable due to the discontinuity in the given interval
0%
Yes Rolle's theorem is applicable and the stationary point is $x= ab$
0%
none of these

Explanation

We have

$\displaystyle f(a)=\log \left [ \frac{a^{2}+ab}{(a+b)a} \right ]=\log 1=0$ and

$\displaystyle f(b)=\log \left [ \frac{b^{2}+ab}{(a+b)b} \right ]=\log1=0$

$\Rightarrow f(a)=f(b)=0.$ Also, it

can be easily seen that

$f(x)$ is continuous on

$[a,b]$ and differentiable on

$[a,b]$ .
Thus all the three conditions of Rolle's

theorem are satisfied. Hence

$\displaystyle f^{'}(x)=0$ for at last one value of x in

$[a,b]$
Now

$\displaystyle f^{'}(x)=0 = \frac{2x}{x^{2}+ab}-\frac{1}{x}=0\Rightarrow \displaystyle 2x^{2}-(x^{2}+ab)=0=x^{2}=ab$ or

$\displaystyle x=\sqrt{ab}$ which is also known as stationary point.

Verify the Rolle's theorem for the function

$\displaystyle f(x)=x^{2}-3x+2$ on the interval[1,2]

0%
No Rolle's theorem is not applicable in the given interval
0%
Yes Rolle's theorem is applicable in the given interval and the stationary point $x=\frac { 5 }{ 4 }$
0%
Yes Rolle's theorem is applicable in the given interval and the stationary point $x=\frac { 3 }{ 2 }$
0%
nnone of these

Explanation

It can be easily seen that

$f(x)=x^{2}-3x+2$ is continuous as differentiable on R (being a polynomial)

$\Rightarrow f(x)$ is continous in (1,2) and differentiable in [1,2]. Also, we have

$f(1)=f(2)=0$ .

Thus,

$f(x)$ satisfies all the conditions of Rolle's theorem in

$[1,2]$

$\Rightarrow \displaystyle \exists$ at least one number, say

$x$ in

$[1,2]$ such that

$\displaystyle f^{'}(c)=0.$ Now,

$\displaystyle f^{'}(x)=2x-3=0\Rightarrow x=\frac{3}{2}$ Since, the root (stationary point)

$\displaystyle x=\frac{3}{2}$ lies in the interval(1,2).

Hence Rolle's theorem is verified.

Differentiate

$\displaystyle x^{\sin^{-1}x}$ w.r.t.

$\displaystyle \sin ^{-1}x.$

0%
$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$
0%
$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$
0%
$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$
0%
$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$

Explanation

Let

$\displaystyle y = x^{\sin^{-1}x}$ and

$z = \displaystyle \sin ^{-1}x\Rightarrow \sin z = x$
we have to find

$\cfrac{dy}{dz}$

$\Rightarrow y = (\sin z)^z$ taking

$\log$ both side

$\log y = z\log \sin z$
Differentiating w.r.t

$z$

$\displaystyle \frac{1}{y}\frac{dy}{dz}=\log\sin z+z\frac{\cos z}{\sin z}$

$\therefore \displaystyle \frac{dy}{dz}=y\left(\log\sin z+z\frac{\cos z}{\sin z}\right)=\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$

$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$ .Find dy/dx

0%
$\sin x(\cot x)^{ \sin x-1 }(-cosec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(-sinx) }$
0%
$\sin x(\cot x)^{ \sin x-1 }(-co\sec ^{ 2 } x)+\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x }$
0%
$\sin x(\cot x)^{ \sin x-1 }(-sec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x+1 } }co\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(sinx) }$
0%
None of these

Explanation

Given

$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$

Let

$u=(\cot x)^{\sin x}$

Taking log on both sides

$\log u=\sin x \log(\cot x)$

Differentiating w.r.t. x, we get,

$\dfrac{1}{u}\dfrac{du}{dx}=\dfrac{\sin x}{\cot x}(-cosec^2 x)+\cos x \log \cot x$

$\Rightarrow \dfrac{du}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x$

Now,

$v=(\tan x)^{\cos x}$

Taking log on both sides

$\log v=\cos x \log(\tan x)$

Differentiating w.r.t. x, we get,

$\dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{\cos x}{\tan x}(\sec^2 x)-\sin x \log \tan x$

$\Rightarrow \dfrac{dv}{dx}=\cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$

Since,

$y=u+v$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$

$\Rightarrow \dfrac{dy}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x+$

$\cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$

Verify Rolle's theorem for

$\displaystyle f(x)=x(x+3)e^{-x/2}$ in

$(-3,0)$

0%
Yes Rolle's theorem is applicable and the stationary point is $x=-2$
0%
Yes Rolle's theorem is applicable and the stationary point is $x=-1$
0%
No Rolle's theorem is not applicable in the given interval
0%
Both A and B

Explanation

We have

$\displaystyle f(x)=x(x+3)e^{-x/2}$

$\displaystyle \therefore f^{'}(x)=(2x+3)e^{-x/2}+(x^{2}+3x)e^{-x/2}\left ( -\frac{1}{2} \right )$

$\displaystyle =e^{-x/2}\left [ 2x+3-\frac{1}{2}[x^{2}+3x] \right ]=-\frac{1}{2}[x^{2}-x-6]e^{-x/2}$

$f^{'}(x)$ exist for every value of x in the interval[-3,0]
Hence,

$f(x)$ is differentiable and hence, continous in the interval

$[-3,0]$
Also, we have

$\displaystyle f(-3)=f(0)=0\Rightarrow$ All the three conditions of Rolle's theorem are satisfied.
So

$\displaystyle f^{'}(x)=0\Rightarrow \frac{1}{2}(x^{2}-x-6)e^{-x/6}=0\Rightarrow x^{2}-x-6=0 \Rightarrow x=3,-2$
Since, the value

$x=-2$ lies in the open interval

$[-3,0]$ the Rolle's theorem is verified.

Verify Rolle's theorem the function

$\displaystyle f(x)=x^{3}-4x$ on

$[-2,2].$ If you think it is applicable in the given interval then find the stationary point ?

0%
Yes Rolle's theorem is applicable and stationary point is $x=\pm \dfrac{2}{\sqrt{3}}$
0%
No Rolle's theorem is not applicable
0%
Yes Rolle's theorem is applicable and $x=2\ or\ -2$
0%
none of these

Explanation

The function

$\displaystyle f(x)=x^{3}-4x$ is a polynomial and so it is continuous and differentiable at all x

$\displaystyle \epsilon$ R.
In particular it is continuous in the closed interval

$[-2,2]$ .
Also

$f(-2)]=0=f(2)$ . Thus ,

$f(x)$ satisfies all three conditions of Rolle's theorem in

$(-2,2)$ .
Therefore , there must exist at least one real number

$'x'$ in the
open interval

$(-2,2)$ for which

$f'(x)=0$
Also

$\displaystyle f'(x)= 3x^{2}-4$
Now

$\displaystyle f^{'}(x)=0$ gives

$3x^{2}-4=0$ or

$x=\pm \dfrac{2}{\sqrt{3}}$ which is also known as stationary point.
Both these value lie in the open interval

$(-2,2)$ and thus the conclusion of Rolle's theorem is verified.

Verify the Rolle's theorem for the function

$\displaystyle f(x)=x^{2}$ in

$(-1,1)$

0%
Yes Rolle's theorem is applicable and the stationary point is $x=\frac { 1 }{ 2 }$
0%
Yes Rolle's theorem is applicable and the stationary point is $x=0$
0%
No Rolle's theorem is not applicable in the given interval
0%
Both A and B

Explanation

$f\left( x \right) ={ x }^{ 2 }$

${ x }^{ 2 }$ is continuous in

$[-1,1]$ since it is quadratic equation

$f'\left( x \right) =2x$

$2x$ is defined in

$(-1,1)$

$\Rightarrow f\left( x \right)$ is differentiable on

$(-1,1)$

$f\left( -1 \right) =f\left( 1 \right) =1$

There exists a

$c,a\le c\le b$ such that

$f'\left( c \right) =0$

$2c=0$

$\Rightarrow c=0$ which lies in

$(-1,1)$

Verify Rolle's theorem for the function

$\displaystyle f(x)=10x-x^{2}$ in the interval [0,10]

0%
Yes Rolle's theorem is applicable and the stationary point is $x=5$
0%
Yes Rolle's theorem is applicable and the stationary point is $x=4$
0%
No Rolle's theorem is not applicable in the given interval
0%
none of these

Explanation

$f\left( x \right) =10x-{ x }^{ 2 }\quad \quad \left[ 0,10 \right]$

$10x-{ x }^{ 2 }$ is continuous in

$\left[ 0,10 \right]$ since it is polynomial function.

$f'\left( x \right) =10-2x$ is defined for all values of x in

$(0,10)$

$\Rightarrow f\left( x \right)$ is differentiable on

$(0,10)$

$f\left( 0 \right) =f\left( 10 \right) =0$

$\therefore$ There exists a

$c,a\le c\le b$

$0\le c\le 10$

Such that

$f'\left( c \right) =0$

$10-2x=0$

$x=5$

$5$ lies in

$\left[ 0,10 \right]$

$y = \displaystyle (\tan x)^{\log x}$ , then

$\cfrac{dy}{dx} =$

0%
$(\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$
0%
$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx$
0%
$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x }$
0%
none of these

Explanation

Let

$y = \displaystyle (\tan x)^{\log x}$

$\log y = \log x .\log \tan x$
Differentiating both side w.r.t

$x$

$\cfrac{1}{y}.\cfrac{dy}{dx} = \cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x)$

$\Rightarrow \cfrac{dy}{dx} = (\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$

Verify Lagrange's mean value for the function

$f(x)=\displaystyle \sin x$ in

$\displaystyle \left [ 0,\frac{\pi }{2} \right ]$

0%
No Lagrange's theorem is not applicable
0%
Yes Lagrange's theorem is applicable and $\displaystyle c =\cos ^{-1}\left ( \frac{2}{\pi } \right )$
0%
Yes Lagrange's theorem is applicable and $\displaystyle c =\sin ^{-1}\left ( \frac{4}{\pi } \right )$
0%
none of these

Explanation

The function

$f\left( x \right)=\sin { x }$ is continuous and differentiable for all

$x\in R$ .

In particular it is continuous in the closed interval

$\displaystyle \left[ 0,\frac { \pi }{ 2 } \right]$ and differentiable in the open interval

$\displaystyle \left( 0,\frac { \pi }{ 2 } \right)$ as is required for the application of lagrange's mean value theorem for it.

By lagrange's mean value theorem, there must exist at least one value of

$'c'$ of

$x$ lying in the open interval

$\displaystyle \left( 0,\frac { \pi }{ 2 } \right)$ such that

$\displaystyle \frac { f\left( \frac { \pi }{ 2 } \right) -f\left( 0 \right) }{ \left( \frac { \pi }{ 2 } \right) -0 } =f'\left( c \right)$ ...(1)

Let us verify it, we have

$\displaystyle f\left( 0 \right) =\sin { \left( 0 \right) } =0;f\left( \frac { \pi }{ 2 } \right) =\sin { \left( \frac { \pi }{ 2 } \right) } =1$

Also

$f'\left( x \right) =\cos { x }$ gives

$f'\left( c \right) =\cos { c }$ .

Putting these values in (1), we have

$\displaystyle \frac { 1-0 }{ \frac { \pi }{ 2 } } =\cos { \left( c \right) \Rightarrow \cos { \left( c \right) } } =\frac { 2 }{ \pi } \Rightarrow c=\cos ^{ -1 }{ \left( \frac { 2 }{ \pi } \right) }$

Since

$\displaystyle 0<\frac { 2 }{ \pi } <1,$ therefore the principal value of

$\displaystyle c=\cos ^{ -1 }{ \left( \frac { 2 }{ \pi } \right) }$ lies in the open interval

$\displaystyle \left( 0,\frac { \pi }{ 2 } \right)$ and so is the required value of

$c$

This verifies lagrange's mean value theorem

Find 'c' of the mean value theorem, if

$\displaystyle f(x)=x(x-1)(x-2);a=0, b=1/2$

0%
$\displaystyle c=\frac{1+\sqrt{21}}{6}$
0%
$\displaystyle c=\frac{1-\sqrt{21}}{6}$
0%
$\displaystyle c=\frac{1+\sqrt{6}}{3}$
0%
$\displaystyle c=\frac{1-\sqrt{6}}{3}$

Explanation

We have

$f\left( a \right) =f\left( 0 \right) =0$ and

$\displaystyle f\left( b \right) =f\left( \frac { 1 }{ 2 } \right) =\frac { 3 }{ 8 }$

$\displaystyle \therefore \frac { f\left( b \right) -f\left( a \right) }{ b-a } =\frac { \frac { 3 }{ 8 } -0 }{ \frac { 1 }{ 2 } -0 } =\frac { 3 }{ 4 }$
Now

$f\left( x \right) ={ x }^{ 3 }-3{ x }^{ 2 }+2x\\ \Rightarrow f'\left( x \right) ={ 3x }^{ 2 }-6x+2\\ \Rightarrow f'\left( c \right) ={ 3c }^{ 2 }-6c+2$
Putting all these value in lagrange's mean value theorem

$\displaystyle \frac { f\left( b \right) -f\left( a \right) }{ b-a } =f'\left( c \right) ,\left( a<c,b \right)$
We get

$\displaystyle \frac { 3 }{ 4 } ={ 3c }^{ 2 }-6c+2\Rightarrow c=1\pm \frac { \sqrt { 21 } }{ 6 }$
Hence

$\displaystyle c=\frac { 1-\sqrt { 21 } }{ 6 }$ lies in the open interval

$\displaystyle \left( 0,\frac { 1 }{ 2 } \right)$
Therefore it is the required value

Discuss the applicapibility of Rolle's Theorem to the function

$\displaystyle f(x)=x^{2/3}$ in (-1,1).

0%
Yes Rolle's theorem is applicable and the stationary point $x=0$
0%
Yes Rolle's theorem is applicable and the stationary point $x=\frac { 1 }{ 2 }$
0%
No Rolle's theorem is not applicable in the given interval
0%
none of these

Explanation

$\displaystyle f\left( x \right)={ x }^{ \frac { 2 }{ 3 } }\Rightarrow f'\left( x \right) =\frac { 2 }{ x } { x }^{ -\frac { 1 }{ 3 } }$

$\displaystyle\therefore \lim _{ x\rightarrow 0 }{ f'\left( x \right) } =\lim _{ x\rightarrow 0 }{ \frac { 2 }{ 3 } { \left( 0+h \right) }^{ -\frac { 1 }{ 3 } } } =\infty$

$\displaystyle Rf'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } \right\} } =\lim _{ h\rightarrow 0 }{ \left\{ \frac { { h }^{ \frac { 2 }{ 3 } }-1 }{ h } \right\} } =+\infty$

$\displaystyle Lf'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ -h } \right\} } =f'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { { \left( -h \right) }^{ \frac { 2 }{ 3 } } }{ h } \right\} } =-\infty$

$\displaystyle \therefore Lf'\left( 0 \right) \neq Rf'\left( 0 \right)$

$\displaystyle\therefore f'\left( 0 \right)$ does not exist showing that

$f'(x)$ does not exists in the open interval

$(-1,1)$
Hence, Rolle's Theorem is not applicable
although

$f(-1)=f(1)=1$ and

$f(x)$ is continuous in the closed interval

$(-1,1)$

$\displaystyle x=2 \cos t-cos 2t, y=2\sin t-\sin 2t,$ find the value of

$dy/dx$ .

0%
$\tan\displaystyle \frac{3t}{2}$
0%
$\tan\displaystyle \frac{-3t}{2}$
0%
$\tan\displaystyle \frac{3t}{4}$
0%
$\tan\displaystyle \frac{-3t}{4}$

Explanation

$\dfrac{dx}{dt}=\dfrac{d}{dt}\left(2\cos \left(t\right)-\cos \left(2t\right)\right)$

$=\dfrac{d}{dt}\left(2\cos \left(t\right)\right)-\dfrac{d}{dt}\left(\cos \left(2t\right)\right)$

$=-2\sin \left(t\right)-\left(-2\sin \left(2t\right)\right)$

$=-2\sin \left(t\right)+2\sin \left(2t\right)$

$\dfrac{dy}{dt}=\dfrac{d}{dt}\left(2\sin \left(t\right)-\sin \left(2t\right)\right)$

$=\dfrac{d}{dt}\left(2\sin \left(t\right)\right)-\frac{d}{dt}\left(\sin \left(2t\right)\right)$

$=2\cos \left(t\right)-\cos \left(2t\right)\cdot \:2$

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx}=\dfrac{cos(t)-cos(2t)}{-sin(t)+sin(2t)}=\dfrac{2sin(\dfrac{3t}{2})sint}{2cos(\dfrac{3t}{2})sin(t)}=tan(\dfrac{3t}{2})$

The function

$f$ is defined, where

$x=2t-\left| t \right|, t\in R$ . Draw the graph of

$f$ for the interval

$-1\le x\le 1$ . Also discuss its continuity and differentiability at

$x=0$

0%
continuous and differentiable at $x=0$
0%
not continuous but differentiable at $x=0$
0%
neither continuous nor differentiable at $x=0$
0%
differentiable but not continuous at $x=0$

Explanation

For

$t\ge 0$

$x=2t-t=t$

$y={t}^{2}+{t}^{2}=2{t}^{2}$
For

$t<0$

$x=3t, y=0$

$0\le x< 1 \Rightarrow 0\le t<1$ [

$\therefore x=t$ ]

$-1\le x<0 \Rightarrow \cfrac{-1}{3}\le t<0$ [

$\therefore x=3t$ ]

$f$ is continuous and differentiable at

$x=0$

The function

$f(x) = \displaystyle \sin ^{-1}(\cos x)is :-$

0%
discontinuous at $x = 0$
0%
continuous at $x = 0$
0%
differentiable at $x = 0$
0%
none of these

Explanation

$f(0-) = \displaystyle \lim_{h\to 0}\sin ^{-1}(\cos -h)=\lim_{h\to 0}\sin ^{-1}(\cos h) = \sin^{-1}1$ , Since

$\cos(-x) = \cos x$
and

$f(0+) = \displaystyle \lim_{h\to 0}\sin ^{-1}(\cos h) = \sin^{-1}1$
Clearly at

$x=0$ , L.H.L

$=$ R.H.L

$=f(0) \Rightarrow f(x)$ is continuous at

$x =0$
Also

$\displaystyle {y}'=-\frac{\sin x}{\sqrt{1-\cos ^{2}x}}=-\frac{\sin x}{\sqrt{\sin ^{2}x}}=-\frac{\sin x}{\left | \sin x \right |}$
So the function is not differentiable at the points where

$\sin x=0$ ,
that is, for

$x=k\pi \left ( k\in I \right )$ . In particular,

$x=0$ .

Examine the origin for continuity and derivability in case of the function

$f$ defined by

$f(x)=x\tan ^{ -1 }{ (1/x) }$ ,

$x\ne 0$ and

$f(0)=0$ .

0%
continuous but not differentiable at $x=0$
0%
continuous and differentiable at $x=0$
0%
not continuous and not differentiable at $x=0$
0%
none of these

Explanation

$f(0^+) =\displaystyle \lim_{h \rightarrow 0} (htan^{-1}(\cfrac{1}{h})) = 0 \times \cfrac{\pi}{2} = 0$

$f(0^-) = \displaystyle \lim_{h \rightarrow 0 }(-h tan^{-1}(\cfrac{1}{-h})) = 0 \times \cfrac{- \pi}{2} = 0$
Hence the function is continuous at

$x = 0.$

$f'(0^+) = \displaystyle \lim_{h \rightarrow 0} \cfrac{htan^{-1}(\cfrac{1}{h})}{h} = \cfrac{\pi}{2}$

$f(0^+) = \displaystyle \lim_{h \rightarrow 0} (- h tan^{-1}(\cfrac{1}{-h})) = \cfrac{-\pi}{2}$
Hence the function is not differentiable at

$x = 0.$

State true or false:

$x=t^2+3t-8, y=2t^2-2t-5$ , then

$\dfrac {dy}{dx}$ at

$(2, -1)$ is

$\dfrac {6}{7}$ .

0%
True
0%
False

Explanation

$\displaystyle x={ t }^{ 2 }+3t-8\Rightarrow \frac { dx }{ dt } =2t+3$

$\displaystyle y={ 2t }^{ 2 }-2t-5\Rightarrow \frac { dy }{ dt } =4t-2$

$\displaystyle \therefore \frac { dy }{ dx } =\frac { 4t-2 }{ 2t+3 }$
At

$\left( 2,-1 \right)$

$2={ t }^{ 2 }+3t-8\Rightarrow { t }^{ 2 }+3t-10=0\\ \Rightarrow { t }^{ 2 }+5t-2t-10=0\\ \Rightarrow \left( t+5 \right) \left( t-2 \right) =0\Rightarrow t=-5,t=2$
and

$-1={ 2t }^{ 2 }-2t-5\Rightarrow { 2t }^{ 2 }-2t-4=0\\ \Rightarrow { t }^{ 2 }-t-2=0\Rightarrow { t }^{ 2 }-2t+t-2=0\\ \Rightarrow t=2,t=-1$

$\therefore t=2$
Hence

$\displaystyle \frac { dy }{ dx } =\frac { 4.2-2 }{ 2.2+3 } =\frac { 6 }{ 7 }$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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