MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 7 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 7
If $$\displaystyle y=\sqrt{\sin x+y}$$ then $$\displaystyle \frac{dy}{dx}$$ equals to
Report Question
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 $$x\geq 0)$$
Report Question
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
Report Question
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
Report Question
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
$$\displaystyle \dfrac { d }{ dx } \left( x^{ \log x } \right) =\dfrac { d }{ dx } \left( { e }^{ \log { x } \log { x } } \right) $$
$$\displaystyle =\dfrac { d }{ dx } \left( { \left( \log { x } \right) }^{ 2 } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$$
$$\displaystyle =2\log { x } \dfrac { d }{ dx } \left( \log { x } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$$
$$\displaystyle ={ e }^{ { \left( \log { x } \right) }^{ 2 } }2\log { x } \left( \dfrac { d }{ dx } \left( \log { x } \right) \right) $$
$$\displaystyle =\dfrac { 1 }{ x } 2\log { x } { e }^{ { \left( \log { x } \right) }^{ 2 } }=2x^{ \log { x } -1 }\log { x } $$
If $$y \sin x = x + y$$ then $$\displaystyle \left ( \frac{dy}{dx} \right )_{x=0}$$ equals
Report Question
0%
$$1$$
0%
$$-1$$
0%
$$0$$
0%
$$2$$
Explanation
Given, $$y \sin x = x + y$$
$$\displaystyle y=\frac{x}{\sin x-1}$$
$$\displaystyle \frac{dy}{dx}=\frac{(\sin x-1).1-x.\cos x}{(\sin x-1)^{2}}$$
$$\displaystyle \left ( \frac{dy}{dx} \right )=\frac{-1}{1}=-1$$
If $$\displaystyle x=y\ ln(xy)$$, then $$\displaystyle \frac{dx}{dy}$$ equals
Report Question
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
$$x=y\ln { \left( xy \right) } $$
$$\displaystyle \Rightarrow \dfrac { dx }{ xy } =\ln { \left( xy \right) } +\dfrac { y\left( \dfrac { dx }{ dy } y+x \right) }{ \left( xy \right) } $$
$$\displaystyle \Rightarrow x\dfrac { dx }{ dy } =x\ln { \left( xy \right) } +y\dfrac { dx }{ dy } +x$$
$$\displaystyle \Rightarrow \dfrac { dx }{ dy } \left( x-y \right) =x\ln { \left( xy \right) } +x$$
$$\displaystyle \Rightarrow \dfrac { dx }{ dy } =\dfrac { x\left( x+y \right) }{ y\left( x-y \right) } $$
If $$\displaystyle g(x)=x\tan ^{-1}x $$ then the value of $$g'(1)$$ equals-
Report Question
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
$$g\left( x \right) =x\tan ^{ -1 }{ x } $$
$$\displaystyle \Rightarrow g'\left( x \right) =\tan ^{ -1 }{ x } +\frac { x }{ 1+{ x }^{ 2 } } $$
$$\displaystyle \Rightarrow g'\left( 1 \right) =\tan ^{ -1 }{ \left( 1 \right) } +\frac { 1 }{ 1+{ \left( 1 \right) }^{ 2 } } =\frac { \pi }{ 4 } +\frac { 1 }{ 2 } $$
If $$\displaystyle x^{y}+y^{x}=1$$ then $$\left ( \displaystyle \frac{dy}{dx} \right )$$ equals
Report Question
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 $$x$$
$$\Rightarrow e^{y\log x}\left(\cfrac{y}{x}+\log x\cfrac{dy}{dx}\right)+e^{x\log y}\left(\log y +\cfrac{x}{y}\cfrac{dy}{dx} \right)=0$$
$$\Rightarrow x^y\left(\cfrac{y}{x}+\log x\cfrac{dy}{dx}\right)+y^x\left(\log y +\cfrac{x}{y}\cfrac{dy}{dx} \right)=0$$
$$\Rightarrow \cfrac{dy}{dx} = \displaystyle -\dfrac{yx^{y-1}+y^{x}\log y}{x^{y}\log x+xy^{x-1}}$$
If $$\displaystyle x^{3}-y^{3}+3xy^{2}-3x^{2}y+1=0$$, then at $$(0 , 1)$$ $$\displaystyle \frac{dy}{dx}$$ equals
Report Question
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, $$\displaystyle \frac{dy}{dx}=\frac{\delta f/\delta x}{\delta f/\delta y}=\frac{3}{-3}=-1$$
If $$\sqrt {\dfrac {x}{y}} + \sqrt {\dfrac {y}{x}} = \sqrt {a}$$ then $$y . \dfrac {dx}{dy} =$$
Report Question
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. $$x$$
$$\dfrac {y - xy^{'}}{y^{2}} + \dfrac {xy^{'} - y}{x^{2}} = 0$$
$$\dfrac {1}{y} - \dfrac {xy^{'}}{y^{2}} = \dfrac {y^{'}}{x} - \dfrac {y}{x^{2}}$$
$$\Rightarrow y^{'} \left [\dfrac {1}{x} + \dfrac {x}{y^{2}} \right ] = \dfrac {1}{y} + \dfrac {y}{x^{2}} $$
$$\Rightarrow y^{'} \dfrac {y^{2}+x^{2}}{xy^{2}} = \dfrac {x^{2} + y^{2}}{x^{2}y} $$
$$\therefore \dfrac {dy}{dx} = \dfrac {y}{x}\Rightarrow y \dfrac {dx}{dy} = x$$
If $$\displaystyle x\sqrt{y}+y\sqrt{x}=1$$, then $$\displaystyle \frac{dy}{dx}$$ equals
Report Question
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
$$x\sqrt { y } +y\sqrt { x } =1$$
$$\displaystyle \Rightarrow \sqrt { y } +\frac { x }{ 2\sqrt { y } } \frac { dy }{ dx } +\frac { dy }{ dx } \sqrt { x } +y\frac { 1 }{ 2\sqrt { x } } =0$$
$$\displaystyle \Rightarrow \left( \frac { x }{ 2\sqrt { y } } +\sqrt { x } \right) \frac { dy }{ dx } =-\frac { y }{ 2\sqrt { x } } -\sqrt { y } $$
$$\displaystyle \Rightarrow \left( \frac { x+2\sqrt { xy } }{ 2\sqrt { y } } \right) \frac { dy }{ dx } =-\frac { y-2\sqrt { xy } }{ 2\sqrt { x } } $$
$$\displaystyle \Rightarrow \frac { dy }{ dx } =-\sqrt { \frac { y }{ x } } \left( \frac { y+2\sqrt { xy } }{ x+2\sqrt { xy } } \right) $$
If $$\displaystyle 2^{x}+2^{y}=2^{x+y}$$ then $$\displaystyle \frac{dy}{dx} $$ is equal to
Report Question
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
$$2^{ x }+2^{ y }=2^{ x+y }$$
$$2^{ x }\log { 2 } +(2^{ y }\log { 2 } )\dfrac { dy }{ dx } =(2^{ x+y }\log { 2 } )\left(1+\dfrac { dy }{ dx } \right)$$
$$2^{ x }+2^{ y }\dfrac { dy }{ dx } =2^{ x+y }+2^{ x+y }\dfrac { dy }{ dx } $$
$$(2^{ y }-2^{ x+y })\dfrac { dy }{ dx } =2^{ x+y }-2^{ x }$$
$$\dfrac { dy }{ dx } =\dfrac { 2^{ x+y }-2^{ x } }{ 2^{ y }-2^{ x+y } } $$
$$\dfrac { dy }{ dx } =\dfrac { 2^{ x }(2^{ y }-1) }{ 2^{ y }(1-2^{ x }) } $$
$$\dfrac { dy }{ dx } =\dfrac { 2^{ x-y }(2^{ y }-1) }{ 1-2^{ x } } $$
If $$y = f(x)$$ be a function satisfying the relation $$y^2- x^2 y = x$$, then which of the following may hold good for $$y =f (x)$$ ?
Report Question
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%
$$f'(1) = 1 - \displaystyle \frac{2}{\sqrt 5}$$
If $$y = sec x^0$$ then $$\displaystyle \frac{dy}{dx} = $$
Report Question
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
Report Question
0%
$$0$$
0%
$$\displaystyle \frac {\pi}{2}$$
0%
$$\displaystyle -\frac {\pi}{2}$$
0%
$$\pi$$
Explanation
$$\displaystyle y=(1+x^{2})\tan^{-1}(x-x)$$
$$\Rightarrow \displaystyle \dfrac {dy}{dx}=2x\tan^{-1}x+(1+x^2).\dfrac{1}{1+x^2}-1=2x\tan^{-1}x$$
$$\displaystyle\Rightarrow f(x)= \dfrac {\dfrac {dy}{dx}}{2x}=\tan^{-1}x$$
$$\therefore \displaystyle f(x)+\cot^{-1}x= \tan^{-1}+\cot^{-1}x=\dfrac {\pi}{2}$$
If $$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$$, then $$y'(0)$$ is equal to
Report Question
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 $$x=0, y=1$$,
$$\therefore \displaystyle \frac {dy}{dx}=1+1+1-1-1=1$$
If $$\displaystyle y\sqrt{1+x}+x\sqrt{1+y}=0$$ then value of $$\displaystyle \frac{dy}{dx}$$ at $$y = 1$$ is,
Report Question
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 $$y = 1 , x = \displaystyle -\frac{1}{2}$$
$$\therefore \displaystyle\left ( \frac{dy}{dx} \right )\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}+\sqrt{2}-\frac{1}{4\sqrt{2}}\frac{dy}{dx}=0$$
$$\Rightarrow \dfrac{dy}{dx}=-4$$
If $$y\displaystyle =\sqrt{x\log _{e}x,}$$ then $$\displaystyle \frac{dy}{dx}$$ at $$x= e$$ is-
Report Question
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
Report Question
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 $$g'(x)=3x^2+1$$
$$\Rightarrow f'(1)+g'(2)=2+13=15$$
If $$x^py^q=(x+y)^{p+q}$$, then $$\dfrac{dy}{dx}$$ is equal to
Report Question
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
Report Question
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
$$\dfrac {1}{y}\dfrac {dy}{dx} = y\times \dfrac {1}{x} + \log x \dfrac {dy}{dx}$$
$$\Rightarrow \dfrac {1}{y}\dfrac {dy}{dx} - \log x \dfrac {dy}{dx} = \dfrac {y}{x}$$
$$\Rightarrow \dfrac {dy}{dx}\left (\dfrac {1}{y} - \log x\right ) = \dfrac {y}{x}$$
$$\Rightarrow \dfrac {dy}{dx}\left (\dfrac {1 - y\log x}{y}\right ) = \dfrac {y}{x}$$
$$\Rightarrow \dfrac {xdy}{dx} = \dfrac {y^{2}}{(1 - y \log x)}$$
If $$f\left( x \right) =\sqrt { 1+\cos ^{ 2 }{ \left( { x }^{ 2 } \right) } } $$, then $$f^{ \prime }\left( \dfrac { \sqrt { \pi } }{ 2 } \right) $$ equal to
Report Question
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
$$f^{ \prime }\left( x \right) =\dfrac { -2\sin { { x }^{ 2 } } \cos { { x }^{ 2 } } }{ 2\sqrt { 1+\cos ^{ 2 }{ \left( { x }^{ 2 } \right) } } } \left( 2x \right) .........\left (\cos^2x=\dfrac{\cos 2x+1}2\right )$$
$$\Rightarrow f^{ \prime }\left( x \right) =\dfrac { -\sin { 2{ x }^{ 2 } } }{ \sqrt { 1+\cos ^{ 2 }{ \left( { x }^{ 2 } \right) } } } \left( x \right) $$
$$\therefore f^{ \prime }\left( \dfrac { \sqrt { \pi } }{ 2 } \right) =-\dfrac { \sqrt { \pi } }{ 2 } \cdot \dfrac { \sin { 2\left( \dfrac { \pi }{ 4 } \right) } }{ \sqrt { 1+\dfrac { 1 }{ 2 } } } =-\sqrt { \dfrac { \pi }{ 6 } } $$
If $$x^{p} + y^{q} = (x + y)^{p + q}$$, then $$\dfrac {dy}{dx}$$ is
Report Question
0%
$$-\dfrac {x}{y}$$
0%
$$\dfrac {x}{y}$$
0%
$$-\dfrac {y}{x}$$
0%
$$\dfrac {y}{x}$$
Explanation
If $$x^{p} + y^{q} = (x + y)^{p + q}$$
Taking log on both sides,
$$p\log x + q\log y = (p + q) \log (x + y)$$
On differentiating w.r.t. x, we get
$$\dfrac {p}{x} + \dfrac {q}{y}\cdot \dfrac {dy}{dx} = \dfrac {(p + q)}{(x + y)} \left (1 + \dfrac {dy}{dx}\right )$$
$$\left \{\dfrac {p}{x} - \dfrac {p + q}{x + y}\right \} = \left \{\dfrac {p + q}{x + y} - \dfrac {q}{y}\right \} \dfrac {dy}{dx}$$
$$\left \{\dfrac {px + py - px - qx}{x (x + y)}\right \} = \left \{\dfrac {py + qy - qx - qy}{y (x + y)}\right \} \dfrac {dy}{dx}$$
$$\Rightarrow \dfrac {(py - qx)}{x} = \dfrac {(py - qx)}{y}\cdot \dfrac {dy}{dx}$$
$$\Rightarrow \dfrac {dy}{dx} = \dfrac {y}{x}$$
If $$y=a\sin ^{ 3 }{ \theta } $$ and $$x=a\cos ^{ 3 }{ \theta } $$, then at $$\theta =\dfrac { \pi }{ 3 } ,\dfrac { dy }{ dx } $$ is equal to:
Report Question
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 $$\dfrac { dx }{ d\theta } =-3a\cos ^{ 2 }{ \theta } \sin { \theta } $$
$$\therefore \dfrac { dy }{ dx } =\dfrac { \dfrac { dy }{ d\theta } }{ \dfrac { dx }{ d\theta } } =-\dfrac { 3a\sin ^{ 2 }{ \theta } \cos { \theta } }{ 3a\cos ^{ 2 }{ \theta } \sin { \theta } } $$
$$=-\dfrac { \sin { \theta } }{ \cos { \theta } } =-\tan { \theta } $$
At $$\theta =\dfrac { \pi }{ 3 } , \dfrac { dy }{ dx } =-\tan { \dfrac { \pi }{ 3 } } =-\sqrt { 3 } $$
If $$f(x) = e^xg(x), g(0)=2, g'(0)=1$$, then $$f'(0)$$ is
Report Question
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
Report Question
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, $$\cfrac { du }{ dv } =\cfrac { du/dx }{ dv/dx } $$
$$=\cfrac { \sin { 2x } }{ -\sin { 2x } } =-1$$
If $$\displaystyle xy=1+\log { y } $$ and $$\displaystyle k.\frac { dy }{ dx } +{ y }^{ 2 }=0$$ then k is
Report Question
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
Report Question
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 $$x^2 + y^2 - 20y + 2x\left[(10 - y) \dfrac{dy}{dx} - x\right] + \left[(10 - y) \dfrac{dy}{dx} - x\right]^2$$
$$= x^2 + y^2 - 20y + 2x(10 - y) \dfrac{dy}{dx} - 2x^2 + (10 - y)^2 \left(\dfrac{dy}{dx}\right)^2 + x^2 - 2x(10 - y) \dfrac{dy}{dx}$$
$$= y^2 - 20y + (10 - y)^2\left(\dfrac{dy}{dx}\right)^2$$
If $$y = (1+x) (1+x^2)(1+x^4)......(1+x^{2n})$$ then the value of $$\begin{pmatrix}\dfrac{dy}{dx}\end{pmatrix}$$ at $$x=0$$ is
Report Question
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 $$x=2$$?
Report Question
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 $$1$$.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
