Tag: differencial calculus - differenciability and methods of differnciation

Questions Related to differencial calculus - differenciability and methods of differnciation

Differentiate $\tan^{-1} \sqrt{\dfrac{1+\cos x}{1- \cos x}}$

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

  1. $\dfrac{-1}{2}$

  2. $\dfrac{1}{4}$

  3. $\dfrac{1}{8}$

  4. None of these

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Correct Option: A
Explanation:

Let $y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 + \cos x}}{{1 - \cos x}}} $

$ = {\tan ^{ - 1}}\sqrt {\dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2{{\sin }^2}\dfrac{x}{2}}}} $
$ = {\tan ^{ - 1}}\cot \left( {\dfrac{x}{2}} \right)$
$ = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right)} \right]$
$ = \dfrac{\pi }{2} - \dfrac{x}{2}$
$ \Rightarrow \dfrac{{dy}}{{dx}} =  - \dfrac{1}{2}$

If $y=\dfrac{1+x^2+x^4}{1+x+x^2}$ and $\dfrac{dy}{dx}=ax+b$, then the values of $a$ and $b$ are,

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

  1. $a=2,b=1$

  2. $a=-2,b=1$

  3. $a=2,b=-1$

  4. $a=-2,b=-1$

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Correct Option: C
Explanation:
$\displaystyle y=\frac{1+x^{2}+x^{4}}{1+x+x^{2}}$

$\displaystyle \frac{dy}{dx}= \frac{(2x+4x^{3})(1+x+x^{2})-(1+2x)(1+x^{2}+x^{4})}{(1+x^{2}+x^{4}+2x+2x^{2}+2x^{3})}$

$\displaystyle \frac{dy}{dx}=\frac{2x+2x^{2}+2x^{3}+4x^{3}+4x^{4}+4x^{5}-1-x^{2}-x^{4}-2x-2x^{3}-2x^{5}}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle \frac{dy}{dx}=\frac{2x^{5}+3x^{4}+4x^{3}+x^{2}-1}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle (\frac{dy}{dx})=\frac{(2x^{5}+4x^{4}+6x^{3}+4x^{2}+2x)-(x^{4}+2x^{3}+3x^{2}+2x+1)}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle (\frac{dy}{dx})=\frac{2x[x^{4}+2x^{3}+3x^{2}+2x+1]}{(x^{4}+2x^{3}+3x^{2}+2x+1)}-1$

$\displaystyle (\frac{dy}{dx})=2x-1= ax+b$

$\Rightarrow a=2$ & $b=-1$

If $x=1(\theta+sin\,\theta), y=a(1-cos \theta)$, then at $\theta=\dfrac{\pi}{2},y'=\dfrac{2}{a}$.

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

  1. True

  2. False

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Correct Option: B

If $f\left( x \right) =|x-2|,g\left( x \right) =f\left( f\left( x \right) \right) $, then for $x>4$, $g'(x)=$

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. 2

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Correct Option: A

Differential coefficient of $\log\ \sin x$ is :

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

  1. $\cos x$

  2. $\tan x$

  3. $\text{cosec} \,x$

  4. $\cot x$

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Correct Option: D
Explanation:

We have,

$y=\log \sin x$

On differentiating w.r.t $x$, we get
$\dfrac{dy}{dx}=\dfrac{d(\log \sin x)}{dx}$
$\dfrac{dy}{dx}=\dfrac{1}{\sin x}\times \cos x$
$\dfrac{dy}{dx}=\dfrac{\cos x}{\sin x}$
$\dfrac{dy}{dx}=\cot x$

Hence, this is the answer.

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

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

  1. $f^{ ' }\left( x \right) =1,\forall x$

  2. $f^{ ' }\left( x \right) =1, \forall x\ge 1$

  3. $f^{ ' }\left( x \right) =1, \forall x\le 1$

  4. $f^{ ' }\left( x \right) =1,if\quad x>1\quad and\quad f^{ ' }\left( x \right) =-1\quad if\quad x<1$

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Correct Option: D

The value of sin $ 2^o $ is approximately

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

  1. $ 2^o $

  2. $0.035$

  3. $ \frac {\pi}{180} $

  4. $0.017$

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Correct Option: B

Derivative of $(\sin x)^x + \sin^{-1} \sqrt{x}$ with respect to $x$ is

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

  1. $(x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$

  2. $(x \cot x + \log \sin x) + \dfrac{1}{\sqrt{x - x^2}}$

  3. $(\sin x)^x (x \cot x + \log \,x) + \dfrac{1}{\sqrt{x - x^2}}$

  4. $(\sin x)^x (x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$

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Correct Option: D
Explanation:

Let $y=(\sin x)^x$

$\Rightarrow \log y=x \log (\sin x)$
Now differentiating both sides with respect to $x$ 
$\dfrac{1}{y}\dfrac{dy}{dx}=x\cot x+\log \sin x$
or, $\dfrac{dy}{dx}=(\sin x )^x{x\cot x+\log \sin x}$........(1).
Again let $z=(\sin x)^x+\sin^{-1}\sqrt{x}$
Now differentiating both sides with respect to $x$.
$\dfrac{dz}{dx}=\dfrac{dy}{dx}+\dfrac{1}{\sqrt{1-x}}.\dfrac{1}{2\sqrt{x}}$
$\dfrac{dz}{dx}=(\sin x)^x{x\cot x+\log \sin x}+\dfrac{1}{2\sqrt{x-x^2}}$ [Using (1)]

Let f(x) be a differentiable function satisfying $f(x+y)=f(x)+f(y)\forall x, y \in R$ and $f(0)=1$ then $\displaystyle\lim _{x\rightarrow 0}\dfrac{2^{f(\tan^2x)}-2^{f(\sin^2x)}}{x^3f(\sin x)}$ equals to?

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

  1. $\dfrac{1}{2} ln2$

  2. $ln 2$

  3. $\dfrac{1}{4}ln 2$

  4. $\dfrac{1}{8} ln2$

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Correct Option: A

If $t={ \sin {  }  }^{ -1 }{ 2 }^{ s }$ Then $\dfrac { ds }{ dt }$ is equal to

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

  1. $\dfrac { \log { 2 } }{ \sqrt { 1-t^{ 2 } } }$

  2. $\dfrac { \sin { t } }{ \log { 2 } }$

  3. $\dfrac { \cot { t } }{ \log { 2 } }$

  4. None of these

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Correct Option: D