Question

If y= tan inverse (√1+x2-1/x) or z= tan inverse (2x/1+x2) then dy/dz?

Answer 1

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

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{ \sqrt{1 + {x}^{2}} - 1 }{x}$

and

$\rm :\longmapsto\:z = {tan}^{ - 1}\dfrac{2x}{1 + {x}^{2} }$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{dz}{dy}$

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

Given that,

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{ \sqrt{1 + {x}^{2}} - 1 }{x}$

$\red{\rm :\longmapsto\:Put \: x \: = tana \: so \: that \: a = {tan}^{ - 1}x}$

So, given function can be rewritten as

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{ \sqrt{1 + {tan}^{2}a} - 1 }{tana}$

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{ \sqrt{{sec}^{2}a} - 1 }{tana}$

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{ seca - 1 }{tana}$

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{ \dfrac{1}{cosa} - 1 }{\dfrac{sina}{cosa} }$

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{ \dfrac{1 - cosa}{cosa}}{\dfrac{sina}{cosa} }$

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{1 - cosa}{sina}$

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{2 {sin}^{2} \dfrac{a}{2} }{2sin\dfrac{a}{2}cos\dfrac{a}{2} }$

$\rm :\longmapsto\:y = {tan}^{ - 1}\dfrac{{sin}\dfrac{a}{2} }{cos\dfrac{a}{2} }$

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

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

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

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

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

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} \times \dfrac{1}{1 + {x}^{2} }$

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

Also, given that,

$\rm :\longmapsto\:z = {tan}^{ - 1}\dfrac{2x}{1 + {x}^{2} }$

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

$\rm :\longmapsto\:\dfrac{d}{dx}z = \dfrac{d}{dx} {tan}^{ - 1}\dfrac{2x}{1 + {x}^{2} }$

$\rm :\longmapsto\:\dfrac{dz}{dx}= \dfrac{1}{1 + {\bigg(\dfrac{2x}{1 + {x}^{2} } \bigg) }^{2} } \dfrac{d}{dx}\dfrac{2x}{1 + {x}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {(1 + {x}^{2}) }^{2} }{ {(1 + {x}^{2}) }^{2} + {4x}^{2} } \times \dfrac{ {1 + x}^{2}\dfrac{d}{dx}2x - 2x\dfrac{d}{dx}(1 + {x}^{2} ) }{ {(1 + {x}^{2}) }^{2} }$

$\rm \:  =  \:  \: \dfrac{ 1}{ {(1 + {x}^{2}) }^{2} + {4x}^{2} } \times \dfrac{ ({1 + x}^{2})2 - 2x(0 + 2x) }{ 1 }$

$\rm \:  =  \:  \: \dfrac{2 + {2x}^{2} - {4x}^{2} }{ {(1 + {x}^{2}) }^{2} + {4x}^{2} }$

$\rm \:  =  \:  \: \dfrac{2 - {2x}^{2} }{ {(1 + {x}^{2}) }^{2} + {4x}^{2} }$

$\bf\implies \:\dfrac{dz}{dx} =  \:  \: \dfrac{2 (1 - {x}^{2} )}{ {(1 + {x}^{2}) }^{2} + {4x}^{2} }$

Now,

$\red{\rm :\longmapsto\:\dfrac{dy}{dz}}$

$\rm \:  =  \:  \: \dfrac{dy}{dx} \div \dfrac{dz}{dx}$

$\rm \:  =  \dfrac{1}{2(1 + {x}^{2} )}   \div \: \dfrac{2 (1 - {x}^{2} )}{ {(1 + {x}^{2}) }^{2} + {4x}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {(1 + {x}^{2}) }^{2} + {4x}^{2} }{4(1 + {x}^{2} )(1 - {x}^{2} )}$

$\rm \:  =  \:  \: \dfrac{ {(1 + {x}^{2}) }^{2} + {4x}^{2} }{4(1 - {x}^{4} )}$

Formula Used :-

$\rm :\longmapsto\: {sec}^{2}x - {tan}^{2}x = 1$

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

$\rm :\longmapsto\:sin2x = 2sinx \: cosx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}\dfrac{u}{v} = \dfrac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} }$

$\rm :\longmapsto\:\dfrac{d}{dx}k = 0$

$\rm :\longmapsto\:\dfrac{d}{dx}x = 1$

$\rm :\longmapsto\:\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}$