Question

Find the derivative of Cot^-1(√1+x^2 -1/x)

Answer 1

Answer:

$\dfrac{dy}{dx} \: = \dfrac{1}{2(1 + {x}^{2}) }$

Step-by-step explanation:

To find ---->

$derivative \: of \: \\ {cot}^{ - 1}( \dfrac{ \sqrt{1 + {x}^{2} } - 1 }{x} )$

Concept used ---->

1)

$1 + {tan}^{2} \alpha = {sec}^{2} \alpha$

2)

$1 - cos \alpha = 2 {sin}^{2} \dfrac{ \alpha }{2}$

3)

$sin \alpha = 2 \: sin \dfrac{ \alpha }{2} \: cos \dfrac{ \alpha }{2}$

4)

${cot}^{ - 1} ( \: cot \alpha \: ) = \alpha$

5)

$\dfrac{d}{dx} {tan}^{1} x = \dfrac{1}{1 + {x}^{2} }$

Solution---->

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

$let \: \: \: x \: = tan \alpha \\ = > \alpha \: = {tan}^{ - 1} x$

$= > y = {cot}^{ - 1}( \dfrac{ \sqrt{1 + {tan}^{2} \alpha } - 1 }{tan \alpha } )$

$= {cot}^{ - 1} ( \dfrac{ \sqrt{sec ^{2} \alpha } - 1}{tan \alpha } )$

$= {cot}^{ - 1} ( \dfrac{sec \alpha - 1}{tan \alpha } )$

$= {cot}^{ - 1} ( \: \dfrac{ \dfrac{1}{cos \alpha } - 1 }{ \dfrac{sin \alpha }{cos \alpha } } \: )$

$= {cot}^{ - 1} ( \: \dfrac{ \dfrac{1 - cos \alpha }{cos \alpha } }{ \dfrac{sin \alpha }{cos \alpha } } \: )$

$= {cot}^{ - 1} ( \dfrac{1 - cos \alpha }{sin \alpha } \: )$

$= {cot}^{ - 1} ( \: \dfrac{1 - cos \alpha }{sin \alpha } \: )$

$= {cot}^{1} ( \: \dfrac{2 {sin}^{2} \dfrac{ \alpha }{2} }{2sin \dfrac{ \alpha }{2}cos \dfrac{ \alpha }{2} } \: )$

$= {cot}^{ - 1} ( \: tan \dfrac{ \alpha }{2} \: )$

$= {cot}^{ - 1} ( \: cot \: ( \dfrac{\pi}{2} \: - \dfrac{ \alpha }{2} \: ) \: )$

$= \dfrac{\pi}{2} \: - \dfrac{ \alpha }{2}$

$y= \dfrac{\pi}{2} - \dfrac{1}{2} \: {tan}^{ - 1} x$

$differentiating \: with \: respect \: to \: x \:$

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

$\dfrac{dy}{dx} \: = 0 \: - \dfrac{1}{2} \: ( \: \dfrac{1}{1 + {x}^{2} } \: )$

$\dfrac{dy}{dx} \: = - \dfrac{1}{2(1 + {x}^{2}) }$

Answer 2

$y=cot−1(x1+x2−1)</p><p>\begin{lgathered}let \: \: \: x \: = tan \alpha \\ = > \alpha \: = {tan}^{ - 1} x\end{lgathered}letx=tanα=>α=tan−1x$

$= > y = {cot}^{ - 1}( \dfrac{ \sqrt{1 + {tan}^{2} \alpha } - 1 }$]${tan\alpha } )=>y=cot−1(tanα1+tan2α−1)</p><p>= {cot}^{ - 1} ( \dfrac{ \sqrt{sec ^{2} \alpha } - 1}{tan \alpha } )=cot−1(tanαsec2α−1)$

= {cot}^{ - 1} ( \dfrac{sec \alpha - 1}{tan \alpha } )=cot−1(tanαsecα−1)

= {cot}^{ - 1} ( \: \dfrac{ \dfrac{1}{cos \alpha } - 1 }{ \dfrac{sin \alpha }{cos \alpha } } \: )=cot−1(cosαsinαcosα1−1)

= {cot}^{ - 1} ( \: \dfrac{ \dfrac{1 - cos \alpha }{cos \alpha } }{ \dfrac{sin \alpha }{cos \alpha } } \: )=cot−1(cosαsinαcosα1−cosα)

= {cot}^{ - 1} ( \dfrac{1 - cos \alpha }{sin \alpha } \: )=cot−1(sinα1−cosα)

= {cot}^{ - 1} ( \: \dfrac{1 - cos \alpha }{sin \alpha } \: )=cot−1(sinα1−cosα)

= {cot}^{1} ( \: \dfrac{2 {sin}^{2} \dfrac{ \alpha }{2} }{2sin \dfrac{ \alpha }{2}cos \dfrac{ \alpha }{2} } \: )=cot1(2sin2αcos2α2sin22α)

= {cot}^{ - 1} ( \: tan \dfrac{ \alpha }{2} \: )=cot−1(tan2α)

= {cot}^{ - 1} ( \: cot \: ( \dfrac{\pi}{2} \: - \dfrac{ \alpha }{2} \: ) \: )=cot−1(cot(2π−2α))

= \dfrac{\pi}{2} \: - \dfrac{ \alpha }{2}=2π−2α

y= \dfrac{\pi}{2} - \dfrac{1}{2} \: {tan}^{ - 1} xy=2π−21tan−1x

differentiating \: with \: respect \: to \: x \:differentiatingwithrespecttox

\dfrac{dy}{dx} = \dfrac{d}{dx} ( \dfrac{\pi}{2} ) \: - \dfrac{d}{dx} ( \dfrac{1}{2} {tan}^{ - 1} x \: )dxdy=dxd(2π)−dxd(21tan−1x)

\dfrac{dy}{dx} \: = 0 \: - \dfrac{1}{2} \: ( \: \dfrac{1}{1 + {x}^{2} } \: )dxdy=0−21(1+x21)

\dfrac{dy}{dx} \: = - \dfrac{1}{2(1 + {x}^{2}) }dx[/tex]