Tag: maths

Questions Related to maths

Given $y = x \sqrt{x^2+1}, \dfrac{dy}{dx}$=

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $ \sqrt{x^2+1}$

  2. $\dfrac{2x^2+1}{ \sqrt{x^2+1}}$

  3. $\dfrac{3x^2+1}{ \sqrt{x^2+1}}$

  4. $\dfrac{3x^2+2}{ \sqrt{x^2+1}}$

Show answer
Correct Option: B
Explanation:
Given,

$y=x\sqrt{x^2+1}$

$\Rightarrow \dfrac{d}{dx}y=\dfrac{d}{dx}x\sqrt{x^2+1}$

$=\dfrac{d}{dx}\left(x\right)\sqrt{x^2+1}+\dfrac{d}{dx}\left(\sqrt{x^2+1}\right)x$

$=1\cdot \sqrt{x^2+1}+\dfrac{x}{\sqrt{x^2+1}}x$

$=\dfrac{2x^2+1}{\sqrt{x^2+1}}$

If $\sin { { y+e }^{ -x\cos { y }  } } =e\quad then\quad \frac { dy }{ dx } \quad at\quad (1,\pi )$ is equal to 

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $\sin { y } $

  2. $-x\cos { y } $

  3. $e$

  4. $\sin { y } -x\cos { y } $

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Correct Option: C
Explanation:
Given,

$\sin y+e^{-x\cos y}=e$

$\cos y \dfrac{dy}{dx}+e^{-x\cos y}\left [ x\sin y \dfrac{dy}{dx}-\cos y \right ]=0$

$\Rightarrow \cos y \dfrac{dy}{dx}+e^{-x\cos y} x\sin y \dfrac{dy}{dx} -\cos ye^{-x\cos y}=0$

$\dfrac{dy}{dx}[\cos y+x\sin y e^{-x\cos y}]=\cos y e^{-x\cos y}$

$\dfrac{dy}{dx}=\dfrac{\cos y e^{-x\cos y}}{\cos y+x\sin y e^{-x\cos y}}$

$\left [ \dfrac{dy}{dx} \right ] _{(1, \pi )}=\dfrac{\cos \pi  e^{-\cos \pi}}{\cos \pi+1 \sin \pi e^{-\cos \pi }}$

$\left [ \dfrac{dy}{dx} \right ] _{(1, \pi )}=\dfrac{-1 \times e}{-1+0 \times e}=e$

If $\displaystyle \cos^{-1}\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )=\log a$ then $\displaystyle \frac{dy}{dx}$ is equal to

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $\dfrac{y}{x}$

  2. $\dfrac{x}{y}$

  3. $\displaystyle\dfrac{ x^{2}}{y^{2}}$

  4. $\displaystyle\dfrac{ y^{2}}{x^{2}}$

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Correct Option: A
Explanation:
$\displaystyle \cos^{-1}\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )=\log a$ 
$\Rightarrow \displaystyle \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\cos \log a=A$ (say)
Putting $u=\dfrac{y}{x}$ and applying componendo and dividendo, we have
$\left ( \dfrac{y}{x} \right )^{2}=u^{2}=\left ( 1-A \right )\left ( 1+A \right )$
$\Rightarrow \dfrac{y}{x}=\sqrt{\left ( 1-A \right )\left ( 1+A \right )}\Rightarrow x\dfrac{dy}{dx}-y=0$
$\Rightarrow $   $\dfrac{dy}{dx}=\dfrac{y}{x}$

If $\displaystyle y=\sec(\tan^{-1}x)$, then $\displaystyle \frac {dy}{dx}$ at $x=1$ is equal to

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $\displaystyle \frac {1}{\sqrt {2}}$

  2. $\displaystyle \frac {1}{2}$

  3. $1$

  4. $\sqrt {2}$

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Correct Option: A
Explanation:

Given, $\displaystyle y=\sec(\tan^{-1}x)=\sqrt {1+x^{2}}$
$\Rightarrow \displaystyle \frac {dy}{dx}=\frac {x}{\sqrt {1+x^{2}}}$
$\displaystyle\therefore  \left.\begin{matrix}\frac {dy}{dx}\end{matrix}\right| _{x=1}=\frac {1}{\sqrt {2}}$

If $y=sec(tan^{-1}x)$, then $\displaystyle\frac{dy}{dx}$ is.

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $\displaystyle\frac{x}{\sqrt{1+x^2}}$

  2. $\displaystyle\frac{-x}{\sqrt{1+x^2}}$

  3. $\displaystyle\frac{x}{\sqrt{1-x^2}}$

  4. None of these

Show answer
Correct Option: A
Explanation:

Given, $y=sec(\tan^{-1}x)$
On differentiating w.r.t. $x,$ we get
$\displaystyle\frac{dy}{dx}=sec(\tan^{-1}x)\cdot \tan(\tan^{-1}x)\frac{1}{1+x^2}$
$=\displaystyle\frac{x}{1+x^2}\sqrt{1+x^2}$
$[\because \tan^{-1} x=sec^{-1}(\sqrt{1+x^2})]$
$=\displaystyle\frac{x}{\sqrt{1+x^2}}$

The differential equation $\dfrac {dy}{dx}=\dfrac {1}{ax+by+c}$ where a,b,c are all non zero real number ,is

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. Linear in $y$

  2. Linear in $x$

  3. linear in both $x$ and $y$

  4. Homogeneous equation

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Correct Option: B
Explanation:

$\dfrac {dx}{dy}= ax+by+c$
$\dfrac {dx}{dy}-a=by+c$
Linear in $x$

If ${ x }^{ 2 }.{ e }^{ y }+2x{ ye }^{ x }+13=0$, then $\dfrac { dy }{ dx }$ is

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $\dfrac { -2x{ e }^{ y-x }-2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$

  2. $\dfrac { 2x{ e }^{ x-y }+2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$

  3. $\dfrac { -2x{ e }^{ x-y }+2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$

  4. $None\ of\ these$

Show answer
Correct Option: A
Explanation:

Differentiating w.r.t. $x$

$2xe^y+x^2e^y\dfrac{dy}{dx}+2[\dfrac{d}{dx}(xy)e^x+xye^x]=0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2[\dfrac{d}{{dx}}(xy){e^x} + xy{e^x}] = 0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2{e^x}[y + x\dfrac{{dy}}{{dx}} + xy] = 0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2y{e^x} + 2x{e^x}\dfrac{{dy}}{{dx}} + 2xy2{e^x} = 0$

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

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

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

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


Find: $\dfrac{d}{{\text dx}}\left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right)  \,\,\,$

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. ${\sec ^2}\dfrac{x}{2}$

  2. $\,\dfrac{1}{2}{\sec ^2}\dfrac{x}{2}\,$

  3. $\,\,2{\sec ^2}\dfrac{x}{2}$

  4. $\,\,3{\sec ^2}\dfrac{x}{2}$

Show answer
Correct Option: B
Explanation:

Let $ y = \dfrac{1- \cos x }{\sin x}$


Formula: $\dfrac{d \left (\dfrac{u} {v}\right)}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}$

differentiate on both sides w.r.t $x$


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

$= \dfrac{\sin ^{2}x -cos x + \cos ^{2}x}{(\sin x)^{2}}$

$= \dfrac{1-\cos x}{(\sin x)^{2}}$

$=  \dfrac{2 \sin^{2}\dfrac{x}{2}}{4\sin^{2}\dfrac{x}{2}\cos^{2}\dfrac{x}{2}}$

$= \dfrac{1}{2}\sec^{2}\left(\dfrac{x}{2}\right)$

Derivative of ${ \log { x }  }^{ \cos { x }  }$ with respect to $x$ is

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. ${ \log { x } }^{ \cos { x } }\left[ \cfrac { \cos { x } }{ x\log { x } } -\sin { x } \log { \left( \log { x } \right) } \right] \quad $

  2. ${ \log { x } }^{ \cos { x } }\left[ \cfrac { \cos { x } }{ x\log { x } } -\cos { x } \log { \left( \log { x } \right) } \right] \quad $

  3. ${ \log { x } }^{ \sin { x } }\left[ \cfrac { \sin { x } }{ x\log { x } } -\sin { x } \log { \left( \log { x } \right) } \right] \quad $

  4. None of these

Show answer
Correct Option: D
Explanation:
Let $y=\log x^{\cos x}$

$\Rightarrow \ y=(\cos x)(\log x)$

$\therefore \dfrac {dy}{dx}=(\cos x) \left (\dfrac {1}{x}\right)+(-\sin x)(\log x)$

$=\dfrac {\cos x}{x}-\sin x\log x$

If $xe^{xy}-y=\sin x$, then $\dfrac {dy}{dx}$ at $x=0$ is

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $0$

  2. $1$

  3. $-1$

  4. $None\ of\ these$

Show answer
Correct Option: C
Explanation:
$x _e^ {xy}-y=\sin x$
Differentiating both sides with respect to $x$,
$x\dfrac {d}{dx} (e^{xy}) -\dfrac {d}{dy}+e^{xy}\dfrac {d}{dx}(x)=\dfrac {d}{dx} (\sin x)$
or, $xe^{xy}\dfrac {d}{dx}(xy)-\dfrac {dy}{dx}+e^{xy}=\cos x$
or, $xe^{xy}y+xe^{xy}\dfrac {dy}{dx}-\dfrac {dy}{dx} +e^{xy} +e^{xy}=\cos x$
or, $\dfrac {dy}{dx} (x e^{xy}-1)=\cos x e^{xy} -xye^{xy}$
$\therefore \ \left. \dfrac {dy}{dx}\right] _{x=0}=\dfrac {\cos x-e^{xy} -xye^{xy}}{xe^{xy}-1}=\dfrac {1-0}{0-1}=-1$