Question

$\footnotesize \bold {If \: \sf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log(\sqrt{1 - x^2}) } \\ \footnotesize \bold{ then \: show \: that} \sf \dfrac{dy}{dx} = \dfrac{cos^{-1}x}{(1-x^2)^{\frac{3}{2}}} \\ \footnotesize \bold { Continuity \: a nd \: Differentiability}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log(\sqrt{1 - x^2})$

We know,

$\green{ \boxed{ \sf{ \: log( {x}^{y} ) = y \: logx}}}$

So, using this we get,

$\rm :\longmapsto\:y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - \dfrac{1}{2} log(1 - x^2)$

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - \dfrac{1}{2} \dfrac{d}{dx}log(1 - x^2)$

We know,

$\green{ \boxed{ \sf{ \:\dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u - u\dfrac{d}{dx}v}{ {v}^{2} }}}} \: and \: \green{ \boxed{ \sf{ \:\dfrac{d}{dx}logx = \frac{1}{x}}}}$

So, using this, we get

$\rm=\dfrac{ \sqrt{1 - {x}^{2}}\dfrac{d}{dx} {xcos}^{ - 1}x - x {cos}^{ - 1} x\dfrac{d}{dx} \sqrt{1 - {x}^{2} } }{1 - {x}^{2} } -\dfrac{1}{2(1 - {x}^{2} )}( - 2x)$

We know,

$\green{ \boxed{ \sf{ \:\dfrac{d}{dx}uv \: = \: u\dfrac{d}{dx}v + v\dfrac{d}{dx}u}}} \: \: and \: \: \red{ \boxed{ \sf{ \:\dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}}$

So, using this, we get

$\rm=\dfrac{ \sqrt{1 - {x}^{2}}\bigg[x\dfrac{d}{dx} {cos}^{ - 1}x + {cos}^{ - 1}x\dfrac{d}{dx}x\bigg] - x {cos}^{ - 1} x\dfrac{1}{2 \sqrt{1 - {x}^{2} } }( - 2x) }{1 - {x}^{2} } + \dfrac{x}{(1 - {x}^{2} )}$

We know,

$\green{ \boxed{ \sf{ \:\dfrac{d}{dx}{cos}^{ - 1}x = - \frac{1}{ \sqrt{1 - {x}^{2} } }}}}$

So, using this we get,

$\rm=\dfrac{ \sqrt{1 - {x}^{2}}\bigg[ - x\dfrac{1}{ \sqrt{1 - {x}^{2} } } + {cos}^{ - 1}x\bigg] + \dfrac{ {x}^{2} {cos}^{ - 1}x}{\sqrt{1 - {x}^{2} } }}{1 - {x}^{2} } + \dfrac{x}{(1 - {x}^{2} )}$

$\rm=\dfrac{ - x + \sqrt{1 - {x}^{2} }{cos}^{ - 1}x + \dfrac{ {x}^{2} {cos}^{ - 1}x}{\sqrt{1 - {x}^{2} } }}{1 - {x}^{2} } + \dfrac{x}{(1 - {x}^{2} )}$

$\rm=\dfrac{ - x + \dfrac{(1 - {x)}^{2}{cos}^{ - 1}x + {x}^{2} {cos}^{ - 1}x}{ \sqrt{1 - {x}^{2} } } }{1 - {x}^{2} } + \dfrac{x}{(1 - {x}^{2} )}$

$\rm=\dfrac{ - x + \dfrac{(1 - {x}^{2}+ {x}^{2} ) \: {cos}^{ - 1}x}{ \sqrt{1 - {x}^{2} } } }{1 - {x}^{2} } + \dfrac{x}{(1 - {x}^{2} )}$

$\rm=\dfrac{ - x + \dfrac{\: {cos}^{ - 1}x}{ \sqrt{1 - {x}^{2} } } }{1 - {x}^{2} } + \dfrac{x}{(1 - {x}^{2} )}$

$\rm=\dfrac{ - x }{1 - {x}^{2} } + \dfrac{{cos}^{ - 1}x}{(1 - {x}^{2} ) \sqrt{1 - {x}^{2} } } + \dfrac{x}{(1 - {x}^{2} )}$

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

$\rm= \dfrac{{cos}^{ - 1}x}{ {\bigg(1 - {x}^{2} \bigg) }^{\dfrac{3}{2} } }$

Hence,

$\bf\implies \:\red{ \boxed{ \sf{ \:\dfrac{dy}{dx}= \dfrac{{cos}^{ - 1}x}{ {\bigg(1 - {x}^{2} \bigg) }^{\dfrac{3}{2} } }}}}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Answer:

$\huge \sf{\underbrace{\underline{Answer: -}}}$

To find :-

• Value of the question.

□Explanation :-

• Refer the attachment for more information.
• It is better to refer in this i cant do integration so , it is perfectly to refer the given attachment.

♧Conclusion :-

• I think i did it better for this question.
• If u got any problem in this answer refer the answer given by @mathsdude.

♧Hope it helps u mate.

♧Thank you .