Tag: methods of differentiation

Questions Related to methods of differentiation

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. $None\ of\ these$

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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, $\dfrac {dy}{dx} (x e^{xy}-1)=\cos x e^{xy} -xye^{xy}$
$\therefore \ \left. \dfrac {dy}{dx}\right] _{x=0}=\dfrac {\cos x-e^{xy} -xye^{xy}}{xe^{xy}-1}=\dfrac {1-0}{0-1}=-1$

$\dfrac {d}{dx}(x^{\ell n x})$ is equal to

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

  1. $2x^{\ell n x-1}\ell n x$

  2. $x^{\ell n x-1}$

  3. $2/3(\ell n x)$

  4. $x^{\ell n x-1}.\ell n x$

Show answer
Correct Option: A
Explanation:

Let $lnx=u$

$\therefore\ x={e}^{u}.$
$\frac { d }{ dx } \left( { x }^{ lmx } \right) =\frac { d }{ dx } \left( { e }^{ u.lnu } \right) $
$=\frac { d }{ dx } \left( { e }^{ { lnu }^{ 2 } } \right) =\frac { d }{ dx } \left( { u }^{ 2 } \right) $
$=2u.\frac { du }{ dx } $
$=2u.\frac { d }{ dx } \left( lnx \right) =\boxed{2lnx\quad { x }^{ lnx-1 }}$

If $y=(tan \, x)^{(tan\, x)^{tan\,x}}, $ then at $x=\dfrac{\pi}{4}, \dfrac{dy}{dx}$ is equal to 

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

  1. 0

  2. 3

  3. 2

  4. None of these

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Correct Option: C
Explanation:
$y={ \left( tanx \right)  }^{ { \left( tanx \right)  }^{ \left( tanx \right)  } }$
We have to find $\dfrac { dy }{ dx } $ at $x=\dfrac { \Pi  }{ 4 } $.
Now consider:
$y={ f\left( x \right)  }^{ g\left( x \right)  }$
$ln\left( y \right) =g\left( x \right) lnf\left( x \right) $
$\Rightarrow \dfrac { 1 }{ y } \dfrac { dy }{ dx } =g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  } +{ g }^{ 1 }\left( x \right) lnf\left( x \right) $
$\Rightarrow \dfrac { dy }{ dx } =y\left[ g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  } +{ g }^{ 1 }\left( x \right) lnf\left( x \right)  \right] $
$\Rightarrow \dfrac { dy }{ dx } ={ f\left( x \right)  }^{ g\left( x \right)  }\left[ { g }^{ 1 }\left( x \right) lnf\left( x \right) +g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  }  \right] $
So, We have
$y={ \left( tanx \right)  }^{ { \left( tanx \right)  }^{ tanx } }$
Where  $f\left( x \right) =tan\left( x \right) $
${ f }^{ 1 }\left( x \right) ={ \sec }^{ 2 }x$
$g\left( x \right) ={ \left( tanx \right)  }^{ tanx }$
${ g }^{ 1 }\left( x \right) ={ \left( tanx \right)  }^{ tanx\left[ { \sec }^{ 2 }xln\left( tan\left( x \right)  \right) +\dfrac { \tan\left( x \right) { \sec }^{ 2 }x }{ tanx }  \right]  }$
Now, $\tan\left( \dfrac { \Pi  }{ 4 }  \right) =1$ and $\sec\left( \Pi /4 \right) =\sqrt { 2 } $.
$f\left( \Pi /4 \right) =1$
${ f }^{ 1 }\left( \Pi /4 \right) =2$
$g\left( \Pi /4 \right) =1$
${ g }^{ 1 }\left( \Pi /4 \right) =1\left( 2ln\left( 1 \right) +2 \right) =2$
$\dfrac { dy }{ dx } $ at $\Pi /4$ is
$\dfrac { dy }{ dx } ={ f\left( \Pi /4 \right)  }^{ g\left( \Pi /4 \right) \left[ { g }^{ 1 }\left( \Pi /4 \right) ln\left( f\left( \Pi /4 \right)  \right) +g\left( \Pi /4 \right) \dfrac { { f }^{ 1 }\left( \Pi /4 \right)  }{ f\left( \Pi /4 \right)  }  \right]  }$
$\Rightarrow \dfrac { dy }{ dx } =1\left[ 2ln\left( 1 \right) +1\times 2 \right] =2$
$\Rightarrow \dfrac { dy }{ dx } =2$.

The solution of the differential equation, $y\,dx + \left( {x + {x^2}y} \right)dy = 0$ is

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

  1. $\log y = cx$

  2. $ \log y- \dfrac{1}{{xy}}  = c$

  3. $\dfrac{1}{{xy}} - \log y = c$

  4. $\dfrac{1}{{xy}} + \log y = c$

Show answer
Correct Option: B
Explanation:

We have,

$ydx+\left( { x+{ x^{ 2 } }y } \right) dy=0 \ ydx=-\left( { x+{ x^{ 2 } }y } \right) dy \ ydx+xdy=-{ x^{ 2 } }ydy \ \dfrac { { ydx+xdy } }{ { { { \left( { xy } \right)  }^{ 2 } } } } =-\dfrac { { dy } }{ y }  \ \dfrac { { d\left( { xy } \right)  } }{ { { { \left( { xy } \right)  }^{ 2 } } } } =-\dfrac { { dy } }{ y } $

On taking integrating both sides

$\begin{array}{l} \dfrac { { -1 } }{ { xy } } =-\log  y+c \\ \log  y-\dfrac { 1 }{ { xy } } =c \end{array}$


Hence, this is the answer.

If $x=a\sin \theta$ and $y=b\cos\theta$, then $\displaystyle\frac{d^2y}{dx^2}$ is 

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

  1. $\displaystyle\frac{a}{b^2}\sec^2\theta$

  2. $\displaystyle\frac{-b}{a}\sec^2\theta$

  3. $\displaystyle\frac{b}{a^2}\sec^3\theta$

  4. $\displaystyle\frac{-b}{a^2}\sec^3\theta$

Show answer
Correct Option: D
Explanation:

Given, $x=a\sin\theta$ and $y=b\cos\theta$
On differentiating w.r.t.$\theta$, we get
$\displaystyle\frac{dx}{d\theta}=a\cos\theta$
and $\displaystyle\frac{dy}{d\theta}=-b\sin \theta$
$\Rightarrow \displaystyle\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=-\frac{b}{a}\tan\theta$
Again, differentiating w.r.t. $x$, we get
$\displaystyle\frac{d^2y}{dx^2}=-\frac{b}{a}sec^2\theta\cdot\frac{d\theta}{dx}$
$\Rightarrow \displaystyle\frac{d^2y}{dx^2}=-\frac{b}{a}sec^2\theta\cdot\frac{1}{a\cos\theta}$
$=-\displaystyle\frac{b}{a^2}sec^3\theta$

The set of all points, where the function $f(x) = \sqrt {1 - e^{-x^{2}}}$ is differentiable, is

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

  1. $(0, \infty)$

  2. $(-\infty, \infty)$

  3. $(-\infty, 0) \cup (0, \infty)$

  4. $(-1, \infty)$

Show answer
Correct Option: C
Explanation:

Given that

$ f(x) = \sqrt{1 - e^{-x^2}}$

By chain rule of differentiation, 

$ f'(x) = \dfrac{1}{2\sqrt{1 - e^{-x^2}} } \; \dfrac{d}{dx}(1 - e^{-x^2})$
           $  = \dfrac{1}{2\sqrt{1 - e^{-x^2}} } \; (-e^{-x^2}) (-2x)$
           $  = \dfrac{x \; e^{-x^2}}{2\sqrt{1 - e^{-x^2}} } $

f(x) is differentiable at a point x if the f'(x) exists at the point.

The denominator of f'(x) vanishes for 
$ 1 - e^{-x^2} = 0 $
i.e for
$ x = 0$

f(x) is not differentiable at x = 0.

Hence, the set of all points at which f(x) is differentiable:
$  (-\infty, 0) \; \cup \; (0, \infty) $

The answer: option C.