Question

Find the derivative of tan^-1{x/1+√(1-x²)} w.r.t. sin[2 cot^-1√(1+x)/√(1-x)]
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Answer 1

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

Let we assume that

$\rm :\longmapsto\:u = {tan}^{ - 1}\bigg(\dfrac{x}{1 + \sqrt{1 - {x}^{2} } } \bigg)$

and

$\rm :\longmapsto\:v = sin \: \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ \sqrt{1 + x} }{ \sqrt{1 - x} } \bigg) \bigg\}$

Consider,

$\rm :\longmapsto\:u = {tan}^{ - 1}\bigg(\dfrac{x}{1 + \sqrt{1 - {x}^{2} } } \bigg)$

Put x = siny, we get

$\rm :\longmapsto\:u = {tan}^{ - 1}\bigg(\dfrac{siny}{1 + \sqrt{1 - {sin}^{2}y } } \bigg)$

We know,

$\red{ \boxed{ \rm{ \: {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this

$\rm :\longmapsto\:u = {tan}^{ - 1}\bigg(\dfrac{siny}{1 + \sqrt{{cos}^{2}y } } \bigg)$

$\rm :\longmapsto\:u = {tan}^{ - 1}\bigg(\dfrac{siny}{1 + {{cos}y } } \bigg)$

We know,

$\red{ \boxed{ \rm{ \: 1 + cos2x = {2cos}^{2}x}}}$

and

$\red{ \boxed{ \rm{ \: 1 - cos2x = {2sin}^{2}x}}}$

So, using this identity, we get

$\rm :\longmapsto\:u = {tan}^{ - 1}\bigg(\dfrac{2sin\dfrac{y}{2} cos\dfrac{y}{2} }{2 {cos}^{2}\dfrac{y}{2} } \bigg)$

$\rm :\longmapsto\:u = {tan}^{ - 1}\bigg(\dfrac{sin\dfrac{y}{2} }{ {cos}\dfrac{y}{2} } \bigg)$

$\rm :\longmapsto\:u = {tan}^{ - 1}tan\dfrac{y}{2}$

$\rm :\longmapsto\:u = \dfrac{y}{2}$

$\rm :\longmapsto\:u = \dfrac{1}{2} {sin}^{ - 1}x$

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

$\rm :\longmapsto\:\dfrac{d}{dx}u =\dfrac{d}{dx} \dfrac{1}{2} {sin}^{ - 1}x$

$\rm :\longmapsto\:\dfrac{du}{dx} = \dfrac{1}{2} \dfrac{d}{dx}{sin}^{ - 1}x$

We know,

$\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {sin}^{ - 1}x = \frac{1}{ \sqrt{1 - {x}^{2} } }}}}$

So, using this, we get

$\bf :\longmapsto\:\dfrac{du}{dx} = \dfrac{1}{2} \dfrac{1}{ \sqrt{1 - {x}^{2} } } - - - (1)$

Consider,

$\rm :\longmapsto\:v = sin \: \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ \sqrt{1 + x} }{ \sqrt{1 - x} } \bigg) \bigg\}$

Put x = cosy, we get

$\rm :\longmapsto\:v = sin \: \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ \sqrt{1 + cosy} }{ \sqrt{1 - cosy} } \bigg) \bigg\}$

We know,

$\red{ \boxed{ \rm{ \: 1 + cos2x = {2cos}^{2}x}}}$

and

$\red{ \boxed{ \rm{ \: 1 - cos2x = {2sin}^{2}x}}}$

So, on using these Identities, we get

$\rm :\longmapsto\:v = sin \: \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ \sqrt{2cos ^{2} \dfrac{y}{2} } }{ \sqrt{2sin ^{2} \dfrac{y}{2} } } \bigg) \bigg\}$

$\rm :\longmapsto\:v = sin \: \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ cos \dfrac{y}{2} }{ sin\dfrac{y}{2} } \bigg) \bigg\}$

$\rm :\longmapsto\:v = sin \: \bigg \{2{cot}^{ - 1}\bigg(cot \dfrac{y}{2} \bigg) \bigg\}$

$\rm :\longmapsto\:v = sin \: \bigg \{2\bigg(\dfrac{y}{2} \bigg) \bigg\}$

$\rm :\longmapsto\:v = siny$

We know,

$\red{ \boxed{ \rm{ \: {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this

$\rm :\longmapsto\:v = \sqrt{1 - {cos}^{2} y}$

$\rm :\longmapsto\:v = \sqrt{1 - {x}^{2}}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}v = \dfrac{d}{dx} \sqrt{1 - {x}^{2}}$

We know,

$\red{ \boxed{ \rm{ \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}}$

So, using this

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

$\rm :\longmapsto\:\dfrac{dv}{dx} = \dfrac{1}{2 \sqrt{1 - {x}^{2} } }( - 2x)$

$\rm :\longmapsto\:\dfrac{dv}{dx} = - \: \dfrac{x}{\sqrt{1 - {x}^{2} } } - - - (2)$

Now, Consider

$\rm :\longmapsto\:\dfrac{du}{dv}$

$\rm \: = \: \: \dfrac{du}{dx} \: \div \: \dfrac{dv}{dx}$

$\rm \: = \: \: \dfrac{1}{2 \sqrt{1 - {x}^{2} } } \: \div \: \dfrac{ - x}{ \sqrt{1 - {x}^{2} } }$

$\rm \: = \: \: - \: \dfrac{1}{2x}$

Additional Information :-

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

$\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {tan}^{ - 1}x = \frac{1}{1 + {x}^{2} }}}}$

$\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {cot}^{ - 1}x = \frac{ - \: 1}{1 + {x}^{2} }}}}$

$\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {sec}^{ - 1}x = \frac{1}{x \sqrt{ {x}^{2} - 1} } }}}$

$\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {cosec}^{ - 1}x = \frac{ - \: 1}{x \sqrt{ {x}^{2} - 1} } }}}$