If $$y = sec x^0$$ then $$\displaystyle \frac{dy}{dx} = $$

$$\displaystyle \frac{\pi}{180} sec x^0 tan x^0$$

$$\displaystyle \frac{180}{\pi} sec x^0 tan x^0$$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 7

$\displaystyle y=\sqrt{\sin x+y}$ then

$\displaystyle \frac{dy}{dx}$ equals to

0%
$\displaystyle \frac{\cos x}{2y-1}$
0%
$\displaystyle \frac{\cos }{1-2y}$
0%
$\displaystyle \frac{\sin x}{1-2y}$
0%
$\displaystyle \frac{\sin x}{2y-1}$

Explanation

Given,

$\displaystyle y=\sqrt{\sin x+y}$

$\implies \displaystyle y^{2}=\sin x+y$

Differentiate w.r.t.

$x$ ,

$\implies \displaystyle 2y\frac{dy}{dx}=\cos x +\frac{dy}{dx}$

$\implies \displaystyle \frac{dy}{dx}=\frac{\cos x }{2y-1}$

$\displaystyle \frac{d}{dx}\left ( \tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right ) \right )$ equals

$\displaystyle ($ for

$x\geq 0)$

0%
$\displaystyle \frac{1}{2\sqrt{x}(1+x)}-\frac{1}{1+x^{2}}$
0%
$\displaystyle \frac{1}{2\sqrt{x}(1+x)}+\frac{1}{1+x^{2}}$
0%
$\displaystyle \frac{1}{1+x}-\frac{1}{1+x^{2}}$
0%
None of these

Explanation

Let,

$\displaystyle y=\tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right )$

$\displaystyle y=\tan ^{-1}(\sqrt{x})-\tan ^{-1}x$

$\therefore \displaystyle \frac{dy}{dx}=\frac{1}{(1+x)}\times \frac{1}{2\sqrt{x}}-\frac{1}{1+x^{2}}$

$\displaystyle \frac{d}{dx}\left ( \tan ^{-1}\left ( \frac{a-x}{1+ax} \right ) \right )$ equals if ax > -1

0%
$\displaystyle \frac{a}{1+x^{2}}$
0%
$\displaystyle \frac{1}{1+x^{2}}$
0%
$\displaystyle- \frac{a}{1+x^{2}}$
0%
$\displaystyle -\frac{1}{1+x^{2}}$

Explanation

Let,

$\displaystyle y=\tan ^{-1}\left ( \frac{a-x}{1+ax} \right )$

$\displaystyle y=\tan ^{-1}(a)-\tan ^{-1}(x)$

$\therefore \displaystyle \frac{dy}{dx}=-\frac{1}{1+x^2}$

$\displaystyle \frac{d}{dx}\left ( x^{\log x} \right )$ is equal to

0%
$\displaystyle 2x^{\log x-1}\log x$
0%
$\displaystyle x^{\log x-1}$
0%
$\dfrac 23 (\log x)$
0%
$\displaystyle x^{\log x-1}.\log x$

Explanation

$\displaystyle \dfrac { d }{ dx } \left( x^{ \log x } \right) =\dfrac { d }{ dx } \left( { e }^{ \log { x } \log { x } } \right)$

$\displaystyle =\dfrac { d }{ dx } \left( { \left( \log { x } \right) }^{ 2 } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$

$\displaystyle =2\log { x } \dfrac { d }{ dx } \left( \log { x } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$

$\displaystyle ={ e }^{ { \left( \log { x } \right) }^{ 2 } }2\log { x } \left( \dfrac { d }{ dx } \left( \log { x } \right) \right)$

$\displaystyle =\dfrac { 1 }{ x } 2\log { x } { e }^{ { \left( \log { x } \right) }^{ 2 } }=2x^{ \log { x } -1 }\log { x }$

$y \sin x = x + y$ then

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

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

Explanation

Given,

$y \sin x = x + y$

$\displaystyle y=\frac{x}{\sin x-1}$

$\displaystyle \frac{dy}{dx}=\frac{(\sin x-1).1-x.\cos x}{(\sin x-1)^{2}}$

$\displaystyle \left ( \frac{dy}{dx} \right )=\frac{-1}{1}=-1$

$\displaystyle x=y\ ln(xy)$ , then

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

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

Explanation

$x=y\ln { \left( xy \right) }$

$\displaystyle \Rightarrow \dfrac { dx }{ xy } =\ln { \left( xy \right) } +\dfrac { y\left( \dfrac { dx }{ dy } y+x \right) }{ \left( xy \right) }$

$\displaystyle \Rightarrow x\dfrac { dx }{ dy } =x\ln { \left( xy \right) } +y\dfrac { dx }{ dy } +x$

$\displaystyle \Rightarrow \dfrac { dx }{ dy } \left( x-y \right) =x\ln { \left( xy \right) } +x$

$\displaystyle \Rightarrow \dfrac { dx }{ dy } =\dfrac { x\left( x+y \right) }{ y\left( x-y \right) }$

$\displaystyle g(x)=x\tan ^{-1}x$ then the value of

$g'(1)$ equals-

0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{\pi }{4}$
0%
$\displaystyle \frac{1}{2}-\frac{\pi }{4}$
0%
$\displaystyle \frac{1}{2}+\frac{\pi }{4}$

Explanation

$g\left( x \right) =x\tan ^{ -1 }{ x }$

$\displaystyle \Rightarrow g'\left( x \right) =\tan ^{ -1 }{ x } +\frac { x }{ 1+{ x }^{ 2 } }$

$\displaystyle \Rightarrow g'\left( 1 \right) =\tan ^{ -1 }{ \left( 1 \right) } +\frac { 1 }{ 1+{ \left( 1 \right) }^{ 2 } } =\frac { \pi }{ 4 } +\frac { 1 }{ 2 }$

$\displaystyle x^{y}+y^{x}=1$ then

$\left ( \displaystyle \frac{dy}{dx} \right )$ equals

0%
$\displaystyle \frac{yx^{y-1}+y^{x}\log y}{x^{y}\log x+xy^{x-1}}$
0%
$\displaystyle -\frac{yx^{y-1}+y^{x}\log y}{x^{y}\log x+xy^{x-1}}$
0%
$\displaystyle- \frac{x^{y}\log x +xy^{x-1}}{yx^{y-1}+y^{x}\log y}$
0%
None of these

Explanation

$x^y+y^x = 1\Rightarrow e^{y\log x}+e^{x\log y} = 1$

Now differentiating w.r.t

$x$

$\Rightarrow e^{y\log x}\left(\cfrac{y}{x}+\log x\cfrac{dy}{dx}\right)+e^{x\log y}\left(\log y +\cfrac{x}{y}\cfrac{dy}{dx} \right)=0$

$\Rightarrow x^y\left(\cfrac{y}{x}+\log x\cfrac{dy}{dx}\right)+y^x\left(\log y +\cfrac{x}{y}\cfrac{dy}{dx} \right)=0$

$\Rightarrow \cfrac{dy}{dx} = \displaystyle -\dfrac{yx^{y-1}+y^{x}\log y}{x^{y}\log x+xy^{x-1}}$

$\displaystyle x^{3}-y^{3}+3xy^{2}-3x^{2}y+1=0$ , then at

$(0 , 1)$

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

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

Explanation

$\displaystyle x^{3}-y^{3}+3xy^{2}-3x^{2}y+1=0$

$\cfrac{dy}{dx}=\cfrac{\frac {\delta f}{\delta x}} {\frac {\delta f}{\delta y}}$
Substitute

$\displaystyle f=x^{3}-y^{3}+3xy^{2}-3x^{2}+1$

$\left( \dfrac { \delta f }{ \delta x } \right) =3x^{ 2 }+3y^{ 2 }-6x$

$\displaystyle { \left( \frac { \delta f }{ \delta x } \right) }_{ { (0,1) } }=3$

Again

$\displaystyle f=x^{3}-y^{3}+3xy^{2}-3x^{2}-3x^{2}y+1$

$\left( \dfrac { \delta f }{ \delta y } \right) =-3y^{ 2 }+6xy-6x$

$\displaystyle \left ( \frac{\delta f}{\delta y} \right )_{(0,1)}=-3$

So,

$\displaystyle \frac{dy}{dx}=\frac{\delta f/\delta x}{\delta f/\delta y}=\frac{3}{-3}=-1$

$\sqrt {\dfrac {x}{y}} + \sqrt {\dfrac {y}{x}} = \sqrt {a}$ then

$y . \dfrac {dx}{dy} =$

0%
$\dfrac {x}{y}$
0%
$\dfrac {y}{x}$
0%
$x$
0%
$0$

Explanation

Squaring the given expression,

$\dfrac {x}{y} + \dfrac {y}{x} +2 = a$
Differentiate w.r.t.

$x$

$\dfrac {y - xy^{'}}{y^{2}} + \dfrac {xy^{'} - y}{x^{2}} = 0$

$\dfrac {1}{y} - \dfrac {xy^{'}}{y^{2}} = \dfrac {y^{'}}{x} - \dfrac {y}{x^{2}}$

$\Rightarrow y^{'} \left [\dfrac {1}{x} + \dfrac {x}{y^{2}} \right ] = \dfrac {1}{y} + \dfrac {y}{x^{2}}$

$\Rightarrow y^{'} \dfrac {y^{2}+x^{2}}{xy^{2}} = \dfrac {x^{2} + y^{2}}{x^{2}y}$

$\therefore \dfrac {dy}{dx} = \dfrac {y}{x}\Rightarrow y \dfrac {dx}{dy} = x$

$\displaystyle x\sqrt{y}+y\sqrt{x}=1$ , then

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

0%
$\displaystyle - \frac{y+2\sqrt{xy}}{x+2\sqrt{xy}}$
0%
$\displaystyle -\sqrt{\frac{x}{y}}\left ( \frac{y+2\sqrt{xy}}{x+2\sqrt{xy}} \right )$
0%
$\displaystyle -\sqrt{\frac{y}{x}}\left ( \frac{y+2\sqrt{xy}}{x+2\sqrt{xy}} \right )$
0%
None of these

Explanation

$x\sqrt { y } +y\sqrt { x } =1$

$\displaystyle \Rightarrow \sqrt { y } +\frac { x }{ 2\sqrt { y } } \frac { dy }{ dx } +\frac { dy }{ dx } \sqrt { x } +y\frac { 1 }{ 2\sqrt { x } } =0$

$\displaystyle \Rightarrow \left( \frac { x }{ 2\sqrt { y } } +\sqrt { x } \right) \frac { dy }{ dx } =-\frac { y }{ 2\sqrt { x } } -\sqrt { y }$

$\displaystyle \Rightarrow \left( \frac { x+2\sqrt { xy } }{ 2\sqrt { y } } \right) \frac { dy }{ dx } =-\frac { y-2\sqrt { xy } }{ 2\sqrt { x } }$

$\displaystyle \Rightarrow \frac { dy }{ dx } =-\sqrt { \frac { y }{ x } } \left( \frac { y+2\sqrt { xy } }{ x+2\sqrt { xy } } \right)$

$\displaystyle 2^{x}+2^{y}=2^{x+y}$ then

$\displaystyle \frac{dy}{dx}$ is equal to

0%
$\displaystyle \frac{2^{x}+2^{y}}{2^{x}-2^{y}}$
0%
$\displaystyle \frac{2^{x}+2^{y}}{1+2^{x+y}}$
0%
$\displaystyle 2^{x-y}\left ( \frac{2^{y}-1}{1-2^{x}} \right )$
0%
$\displaystyle \frac{2^{x}+y-2^{x}}{2^{y}}$

Explanation

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

$2^{ x }\log { 2 } +(2^{ y }\log { 2 } )\dfrac { dy }{ dx } =(2^{ x+y }\log { 2 } )\left(1+\dfrac { dy }{ dx } \right)$

$2^{ x }+2^{ y }\dfrac { dy }{ dx } =2^{ x+y }+2^{ x+y }\dfrac { dy }{ dx }$

$(2^{ y }-2^{ x+y })\dfrac { dy }{ dx } =2^{ x+y }-2^{ x }$

$\dfrac { dy }{ dx } =\dfrac { 2^{ x+y }-2^{ x } }{ 2^{ y }-2^{ x+y } }$

$\dfrac { dy }{ dx } =\dfrac { 2^{ x }(2^{ y }-1) }{ 2^{ y }(1-2^{ x }) }$

$\dfrac { dy }{ dx } =\dfrac { 2^{ x-y }(2^{ y }-1) }{ 1-2^{ x } }$

$y = f(x)$ be a function satisfying the relation

$y^2- x^2 y = x$ , then which of the following may hold good for

$y =f (x)$ ?

0%
$f'(x) = \displaystyle \frac{1+2x f(x)}{2f (x) - x^2}$
0%
$f'(x) = \displaystyle \frac{f(x) + 2x f^2 (x)}{f^2 (x) + x}$
0%
$f'(1) = 1 + \displaystyle \frac{2}{\sqrt 5}$
0%
$f'(1) = 1 - \displaystyle \frac{2}{\sqrt 5}$

$y = sec x^0$ then

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

0%
$sec x tan x$
0%
$sec x^0 tan x^0$
0%
$\displaystyle \frac{\pi}{180} sec x^0 tan x^0$
0%
$\displaystyle \frac{180}{\pi} sec x^0 tan x^0$

Explanation

Give

$y = sec x^0$

$= \displaystyle sec \frac{x \pi}{180^o} \left ( \because x^o = \displaystyle \frac{x \pi}{180^o} radians \right )$

$\therefore \displaystyle \frac{dy}{dx} = \frac{\pi}{180^o} sec x^o tan x^o$

Let

$\displaystyle y=(1+x^{2})\tan^{-1}(x-x)$ and

$\displaystyle f(x)=\frac1{2x}\frac {dy}{dx},$ then

$f(x)+\cot^{-1}x$ is equal to

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

Explanation

$\displaystyle y=(1+x^{2})\tan^{-1}(x-x)$

$\Rightarrow \displaystyle \dfrac {dy}{dx}=2x\tan^{-1}x+(1+x^2).\dfrac{1}{1+x^2}-1=2x\tan^{-1}x$

$\displaystyle\Rightarrow f(x)= \dfrac {\dfrac {dy}{dx}}{2x}=\tan^{-1}x$

$\therefore \displaystyle f(x)+\cot^{-1}x= \tan^{-1}+\cot^{-1}x=\dfrac {\pi}{2}$

$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$ , then

$y'(0)$ is equal to

0%
$-1$
0%
$1$
0%
$2$
0%
Non existant

Explanation

$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$
Take log both sides,

$\displaystyle \log y = \frac{1}{3}\log(1+3x)+\frac{1}{4}\log(1+4x)+\frac{1}{5}\log(1+5x)-\frac{1}{7}\log(1+7x)-\frac{1}{8}\log(1+8x)$
Differentiating both side w.r.t

$x$

$\displaystyle \frac {1}{y}\times \frac {dy}{dx}=\frac {1}{(1+3x)}+\frac {1}{1+4x}+\frac {1}{1+5x}-\frac {1}{1+7x}-\frac {1}{1+8x}$
at

$x=0, y=1$ ,

$\therefore \displaystyle \frac {dy}{dx}=1+1+1-1-1=1$

$\displaystyle y\sqrt{1+x}+x\sqrt{1+y}=0$ then value of

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

$y = 1$ is,

0%
$\displaystyle -\frac{1}{2}$
0%
$1$
0%
$-4$
0%
$2$

Explanation

$\displaystyle y\sqrt{1+x}+x\sqrt{1+y}=0$ Given
Differentiating w.r.t

$x$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )\sqrt{1+x}+\frac{1}{2\sqrt{1+x}}y+\sqrt{1+y}+\frac{x}{2\sqrt{1+y}}\frac{dy}{dx}=0$
at

$y = 1 , x = \displaystyle -\frac{1}{2}$

$\therefore \displaystyle\left ( \frac{dy}{dx} \right )\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}+\sqrt{2}-\frac{1}{4\sqrt{2}}\frac{dy}{dx}=0$

$\Rightarrow \dfrac{dy}{dx}=-4$

$y\displaystyle =\sqrt{x\log _{e}x,}$ then

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

$x= e$ is-

0%
$\displaystyle \frac{1}{e}$
0%
$\displaystyle \frac{1}{\sqrt{e}}$
0%
$\displaystyle \sqrt{e}$
0%
None of these

Explanation

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

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac { \dfrac { d }{ dx } \left( x\log { x } \right) }{ 2\sqrt { x\log { x } } }$

$\displaystyle =\dfrac { \log { x } +x\dfrac { 1 }{ x } }{ 2\sqrt { x\log { x } } } =\dfrac { \log { x } +1 }{ 2\sqrt { x\log { x } } }$
At

$x=e$

$\displaystyle \dfrac { dy }{ dx } =\dfrac { 1 }{ \sqrt { e } }$

$\displaystyle f'(1) + g'(2)$ is equal to

0%
$15$
0%
$14$
0%
$13$
0%
$12$

Explanation

Consider

$\displaystyle 6\int f(x)g(x)dx=x^{ 6 }+3x^{ 4 }+3x^{ 2 }+c$

$\Rightarrow \displaystyle \int f(x)g(x)dx=\dfrac { 1 }{ 6 } (x^{ 6 }+3x^{ 4 }+3x^{ 2 }+c)$

Differentiate w.r.t

$x$ ,

$\Rightarrow f(x)g(x)=x^{ 5 }+2x^{ 3 }+x$ ...........

$(1)$

Consider,

$\displaystyle 2\int \dfrac { g(x)dx }{ f(x) } =x^{ 2 }+c$

$\Rightarrow \displaystyle \int \dfrac { g(x)dx }{ f(x) } =\frac { 1 }{ 2 } (x^{ 2 }+c)$

Differentiate w.r.t

$x$

$\Rightarrow \dfrac { g(x) }{ f(x) } =x$ ............

$(2)$

$(1)\times (2)\Rightarrow ({ g(x)) }^{ 2 }=x^{ 6 }+2x^{ 4 }+x^{ 2 }=({ x }^{ 3 }+x)^{ 2 }$

$\Rightarrow g(x)={ x }^{ 3 }+x$

$\dfrac { (1) }{ (2) } \Rightarrow (f(x))^{ 2 }=x^{ 4 }+2x^{ 2 }+1=({ x }^{ 2 }+1)^{ 2 }$

$\Rightarrow f(x)={ x }^{ 2 }+1$

Now,

$f'(x)=2x$ and

$g'(x)=3x^2+1$

$\Rightarrow f'(1)+g'(2)=2+13=15$

$x^py^q=(x+y)^{p+q}$ , then

$\dfrac{dy}{dx}$ is equal to

0%
$\dfrac {y}{x}$
0%
$\dfrac {py}{qx}$
0%
$\dfrac {x}{y}$
0%
$\dfrac {qy}{px}$

Explanation

Given,

$x^py^q=(x+y)^{p+q}$
Taking

$\log$ on both sides, we get

$p\,\log\,x+q\,\log\,y=(p+q)\log(x+y)$

$\Rightarrow \displaystyle \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}= \frac{(p+q)}{(x+y)}\left(1+\frac{dy}{dx}\right)$

$\Rightarrow \displaystyle \left(\frac{p}{x}-\frac{p+q}{x+y}\right)=\left(\frac{p+q}{x+y}-\frac{q}{y}\right)\frac{dy}{dx}$

$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}$

Let

$y$ be the solution of the differential equation

$x \dfrac {dy}{dx} = \dfrac {y^{2}}{1 - y\log x}$ satisfying

$y(1) = 1$ . Then

$y$ satisfies

0%
$y = x^{y - 1}$
0%
$y = x^{y}$
0%
$y = x^{y + 1}$
0%
$y = x^{y + 2}$

Explanation

Let

$y = x^{y}\Rightarrow \log y = y \log x$
Differentiating both sides w.r.t.

$x$ , we get

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

$\Rightarrow \dfrac {1}{y}\dfrac {dy}{dx} - \log x \dfrac {dy}{dx} = \dfrac {y}{x}$

$\Rightarrow \dfrac {dy}{dx}\left (\dfrac {1}{y} - \log x\right ) = \dfrac {y}{x}$

$\Rightarrow \dfrac {dy}{dx}\left (\dfrac {1 - y\log x}{y}\right ) = \dfrac {y}{x}$

$\Rightarrow \dfrac {xdy}{dx} = \dfrac {y^{2}}{(1 - y \log x)}$

$f\left( x \right) =\sqrt { 1+\cos ^{ 2 }{ \left( { x }^{ 2 } \right) } }$ , then

$f^{ \prime }\left( \dfrac { \sqrt { \pi } }{ 2 } \right)$ equal to

0%
$\dfrac { \sqrt { \pi } }{ 6 }$
0%
$-\sqrt { \dfrac { \pi }{ 6 } }$
0%
$\dfrac { 1 }{ \sqrt { 6 } }$
0%
$\dfrac { \pi }{ \sqrt { 6 } }$

Explanation

Given,

$f\left( x \right) =\sqrt { 1+\cos ^{ 2 }{ \left( { x }^{ 2 } \right) } }$ ......(i)

On differentiating equation (i) with respect to

$x$ , we get

$f^{ \prime }\left( x \right) =\dfrac { -2\sin { { x }^{ 2 } } \cos { { x }^{ 2 } } }{ 2\sqrt { 1+\cos ^{ 2 }{ \left( { x }^{ 2 } \right) } } } \left( 2x \right) .........\left (\cos^2x=\dfrac{\cos 2x+1}2\right )$

$\Rightarrow f^{ \prime }\left( x \right) =\dfrac { -\sin { 2{ x }^{ 2 } } }{ \sqrt { 1+\cos ^{ 2 }{ \left( { x }^{ 2 } \right) } } } \left( x \right)$

$\therefore f^{ \prime }\left( \dfrac { \sqrt { \pi } }{ 2 } \right) =-\dfrac { \sqrt { \pi } }{ 2 } \cdot \dfrac { \sin { 2\left( \dfrac { \pi }{ 4 } \right) } }{ \sqrt { 1+\dfrac { 1 }{ 2 } } } =-\sqrt { \dfrac { \pi }{ 6 } }$

$x^{p} + y^{q} = (x + y)^{p + q}$ , then

$\dfrac {dy}{dx}$ is

0%
$-\dfrac {x}{y}$
0%
$\dfrac {x}{y}$
0%
$-\dfrac {y}{x}$
0%
$\dfrac {y}{x}$

Explanation

$x^{p} + y^{q} = (x + y)^{p + q}$

Taking log on both sides,

$p\log x + q\log y = (p + q) \log (x + y)$

On differentiating w.r.t. x, we get

$\dfrac {p}{x} + \dfrac {q}{y}\cdot \dfrac {dy}{dx} = \dfrac {(p + q)}{(x + y)} \left (1 + \dfrac {dy}{dx}\right )$

$\left \{\dfrac {p}{x} - \dfrac {p + q}{x + y}\right \} = \left \{\dfrac {p + q}{x + y} - \dfrac {q}{y}\right \} \dfrac {dy}{dx}$

$\left \{\dfrac {px + py - px - qx}{x (x + y)}\right \} = \left \{\dfrac {py + qy - qx - qy}{y (x + y)}\right \} \dfrac {dy}{dx}$

$\Rightarrow \dfrac {(py - qx)}{x} = \dfrac {(py - qx)}{y}\cdot \dfrac {dy}{dx}$

$\Rightarrow \dfrac {dy}{dx} = \dfrac {y}{x}$

$y=a\sin ^{ 3 }{ \theta }$ and

$x=a\cos ^{ 3 }{ \theta }$ , then at

$\theta =\dfrac { \pi }{ 3 } ,\dfrac { dy }{ dx }$ is equal to:

0%
$\dfrac { 1 }{ \sqrt { 3 } }$
0%
$-\sqrt { 3 }$
0%
$\dfrac { -1 }{ \sqrt { 3 } }$
0%
$\sqrt { 3 }$

Explanation

Given,

$y=a\sin ^{ 3 }{ \theta }$ and

$x=a\cos ^{ 3 }{ \theta }$

On differentiating with respect to

$\theta$ , we get

$\dfrac { dy }{ d\theta } =3a\sin ^{ 2 }{ \theta } \cos { \theta }$

and

$\dfrac { dx }{ d\theta } =-3a\cos ^{ 2 }{ \theta } \sin { \theta }$

$\therefore \dfrac { dy }{ dx } =\dfrac { \dfrac { dy }{ d\theta } }{ \dfrac { dx }{ d\theta } } =-\dfrac { 3a\sin ^{ 2 }{ \theta } \cos { \theta } }{ 3a\cos ^{ 2 }{ \theta } \sin { \theta } }$

$=-\dfrac { \sin { \theta } }{ \cos { \theta } } =-\tan { \theta }$

$\theta =\dfrac { \pi }{ 3 } , \dfrac { dy }{ dx } =-\tan { \dfrac { \pi }{ 3 } } =-\sqrt { 3 }$

$f(x) = e^xg(x), g(0)=2, g'(0)=1$ , then

$f'(0)$ is

0%
$1$
0%
$3$
0%
$2$
0%
$0$

Explanation

$f'(x) = e^xg'(x)+e^xg(x)$

$\Rightarrow f'(0)=e^0.g'(0)+e^0g(0)$

$=1.1+1.2$

$[\because \{g'(0)=1$ and

$g(0)=2\}]$

$=1+2=3$

The derivative of

$\sin ^{ 2 }{ x }$ with respect to

$\cos ^{ 2 }{ x }$ is

0%
$\tan ^{ 2 }{ x }$
0%
$\tan { x }$
0%
$-\tan { x }$
0%
None of these

Explanation

Let

$u=\sin ^{ 2 }{ x } ;\quad v=\cos ^{ 2 }{ x }$

On differentiating w.r.t

$x$ respectively, we get

$\cfrac { du }{ dx } =2\sin { x } \cos { x } =\sin { 2x }$

$\cfrac { dv }{ dx } =-2\cos { x } \sin { x } =-\sin { 2x }$

Now,

$\cfrac { du }{ dv } =\cfrac { du/dx }{ dv/dx }$

$=\cfrac { \sin { 2x } }{ -\sin { 2x } } =-1$

$\displaystyle xy=1+\log { y }$ and

$\displaystyle k.\frac { dy }{ dx } +{ y }^{ 2 }=0$ then k is

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

Explanation

$xy = 1 + \log y$

Differentiating both sides, we get

$x \dfrac{dy}{dx} + y = \dfrac{1}{y} \times \dfrac{dy}{dx}$

Multiplying by y throughout, we get

$xy \dfrac{dy}{dx} + y^2 = \dfrac{dy}{dx}$

$\Rightarrow \dfrac{dy}{dx} (xy - 1) + y^2 = 0$

$\Rightarrow k = xy - 1$

The differential equation of family of circles having centre on line

$y = 10$ and touching x-axis is

0%
$\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -5\frac { dy }{ dx } +6y=0$
0%
$\displaystyle { x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +x\frac { dy }{ dx } +y=0$
0%
$\displaystyle { 8\left( \frac { dy }{ dx } \right) }^{ 3 }-27y=0$
0%
$\displaystyle { \left( y-10 \right) }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ y }^{ 2 }-20y=0$

Explanation

The general equation for a circle can be written as

$x^2 + y^2 + 2gx + 2fy + c = 0$

The center lies on

$y = 10$ , and so

$f = -10.$

Also, distance of the center from x axis has to be equal to the radius.

$\Rightarrow f^2 = g^2 + f^2 - c$

$\Rightarrow g^2 = c$

The equation thus becomes

$x^2 + y^2 + 2gx - 20y + g^2 = 0$

Differentiating w.r.t. x, we get

$2x + 2y \dfrac{dy}{dx} + 2g - 20 \dfrac{dy}{dx} = 0$

$\Rightarrow g = (10 - y) \dfrac{dy}{dx} - x$

Resubstituting this value in the above equation, we get

$x^2 + y^2 - 20y + 2x\left[(10 - y) \dfrac{dy}{dx} - x\right] + \left[(10 - y) \dfrac{dy}{dx} - x\right]^2$

$= x^2 + y^2 - 20y + 2x(10 - y) \dfrac{dy}{dx} - 2x^2 + (10 - y)^2 \left(\dfrac{dy}{dx}\right)^2 + x^2 - 2x(10 - y) \dfrac{dy}{dx}$

$= y^2 - 20y + (10 - y)^2\left(\dfrac{dy}{dx}\right)^2$

$y = (1+x) (1+x^2)(1+x^4)......(1+x^{2n})$ then the value of

$\begin{pmatrix}\dfrac{dy}{dx}\end{pmatrix}$ at

$x=0$ is

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

Explanation

We have,

$y=(1+x)(1+x^2)(1+x^4).......(1+x^4)$

take natural logarithm both sides

$\ln y=\ln(1+x)+\ln(1+x^2)+\ln(1+x^4)+................+\ln(1+x^{2n})$

Now differentiate both sides w.r.t.

$x$

$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{1+x}+\dfrac{2x}{1+x^2}+\dfrac{4x^3}{1+x^4}+......+\dfrac{2nx^{2n-1}}{1+x^{2n}}$

$\Rightarrow \dfrac{dy}{dx}=y\left(\dfrac{1}{x}+\dfrac{2x}{1+x^2}+...........+\dfrac{2nx^{2n-1}}{1+x^{2n}} \right)$

Now substitute

$x=0$ to get the required value

$\Rightarrow \dfrac{dy}{dx}\bigg|_{x=0}=1(1+0+0+......+0)=1$

Note that at

$x=0$ , value of

$y$ is also

$1$

What is the derivative of

$\left| x-1 \right|$ at

$x=2$ ?

0%
$-1$
0%
$0$
0%
$1$
0%
Derivative does not exist

Explanation

given

$f(x)=|x-1|$

$f(x)=(x-1)$ when

$x-1>0$ i.e.

$x>1$

$f(x)=-(x-1)=1-x$ when

$x-1<0$ i.e.

$x<1$

So, at

$x=2, f(x)=x-1$

$\dfrac { df }{ dx } =1$

Thus, derivative of

$f$ at

$x=2$ is

$1$ .

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 7 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 7 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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