Math, asked by Anonymous, 9 hours ago

Solve the attachment!!

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Answered by mathdude500
15

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

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

Let assume that

\rm :\longmapsto\:y =  {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

We know,

\boxed{\tt{  {cosec}^{ - 1}x =  {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}

So, using this, we get

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 +  {x}^{2} } \bigg)

Now, we use Method of Substitution, So we substitute

 \red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z =  {tan}^{ - 1}x}

So, above expression can be rewritten as

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 +  {tan}^{2} z} \bigg)

\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)

\rm\implies \:y = 2z

\bf\implies \:y = 2 {tan}^{ - 1}x

So,

\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x

Thus,

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

\rm \:  =  \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)

\rm \:  =  \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)

\rm \:  =  \: 2 \times \dfrac{1}{1 +  {x}^{2} }

\rm \:  =  \: \dfrac{2}{1 +  {x}^{2} }

Hence,

 \purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) =  \frac{2}{1 +  {x}^{2} }}}}

Hence, Option (d) is correct

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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}

Answered by Anonymous
38

Given :-

 \quad \leadsto \quad \sf \dfrac{d}{dx} \bigg\{ \cosec^{-1} \bigg( \dfrac{1+x²}{2x} \bigg) \bigg\}

To Find :-

The solution of the given from the above options .

Solution :-

Let us assume that ;

 \quad \leadsto \quad \sf f(x) =  \cosec^{-1} \bigg( \dfrac{1+x²}{2x} \bigg)

Now we knows that ;

 \quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \cosec^{-1} \theta = \sin^{-1} \bigg(\dfrac{1}{\theta} \bigg) }}}}}}{\bigstar}

Using this we have ;

 { : \implies \quad \sf f(x) =  \sin^{-1} \bigg( \dfrac{2x}{1+x²} \bigg)}

Now , Put :-

  •  \sf x = \tan \theta

In order to obtain :

  •  \sf \theta = \tan^{-1} x

Now , in the function put the value that we have assumed :

 { : \implies \quad \sf f(x) =  \sin^{-1} \bigg( \dfrac{2 \tan \theta }{1+ \tan² \theta} \bigg)}

Now we knows a Trigonometric identity ;

 \quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \sin 2\theta = \dfrac{2 \tan \theta }{1+ \tan² \theta}}}}}}}{\bigstar}

Now using this we have ;

 { : \implies \quad \sf f(x) =  \sin^{-1} ( \sin 2\theta) }

 { : \implies \quad \sf f(x) =  \cancel{\sin^{-1}} ( \cancel{\sin} 2\theta) }

 { : \implies \quad \sf f(x) =  2\theta }

Now , put the value of  \theta

 { : \implies \quad \sf f(x) =  2 \tan^{-1} x }

Now , differentiate both sides w.r.t.x ;

 { : \implies \quad \sf f'(x) =  \dfrac{d}{dx} ( 2 \tan^{-1} x ) }

Now we knows that ;

 \quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \dfrac{d}{dx} \bigg\{ C . f ( x ) \bigg\}= C . \dfrac{d}{dx} \bigg\{ f ( x ) \bigg\} }}}}}}{\bigstar}

  • Where ' C ' is a constant .

Using this we have ;

 { : \implies \quad \sf f'(x) =  2. \dfrac{d}{dx} (  \tan^{-1} x ) }

Now , we knows that ;

 \quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \dfrac{d}{dx} ( \tan^{-1} x ) = \dfrac{1}{1+x²} }}}}}}{\bigstar}

Using this we have ;

 { : \implies \quad \sf f'(x) =  2 \bigg( \dfrac{1}{1+x²} \bigg) }

 { : \implies \quad \bf f'(x) =  \bigg( \dfrac{2}{1+x²} \bigg) }

  { \bigstar { \underline { \boxed { \red { \bf \therefore \dfrac{d}{dx} \bigg\{ \cosec^{-1} \bigg( \dfrac{1+x²}{2x} \bigg) \bigg\} = \bigg( \dfrac{2}{1+x²} \bigg) }}}}}{\bigstar}

As this matches with none of the given options .

Henceforth , The Required Answer is ( d )

Hope it helps (◕દ◕)

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