Question

if u =sin^ -1 (2x/1+x^2) and v= tan ^-1 (2x/1-x^2) find du/dv
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Answer 1

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

$\displaystyle \sf{ \frac{du}{dv} \: \: when \: \: u \: = { \sin}^{ - 1} \frac{2x}{1 + {x}^{2} } \: \: \: and \: \: \: v = \: \: { \tan}^{ -1 } \frac{2x}{1 - {x}^{2} }}$

CALCULATION

$\sf{ Let \: \: \: x = \tan \theta \: }$

Then

$\displaystyle \sf{ u \: = { \sin}^{ - 1} \frac{2x}{1 + {x}^{2} } \: \: \: gives}$

$\displaystyle \sf{ u \: = { \sin}^{ - 1} \: \: \frac{2 \tan \theta}{1 + { \tan \: }^{2}\theta } \: \: \: }$

$\implies \: \displaystyle \sf{ u \: = { \sin}^{ - 1} ( \sin 2 \theta) \: \: \: }$

$\implies \: \displaystyle \sf{ u \: = 2 \theta \: \: \: }$

Again

$\displaystyle \sf{ v \: = { \tan}^{ - 1} \frac{2x}{1 - {x}^{2} } \: \: \: gives}$

$\displaystyle \sf{ v \: = { \tan}^{ - 1} \: \: \frac{2 \tan \theta}{1 - { \tan \: }^{2}\theta } \: \: \: }$

$\implies \: \displaystyle \sf{ v\: = { \tan}^{ - 1} ( \tan 2 \theta) \: \: \: }$

$\implies \: \displaystyle \sf{ v \: = 2 \theta \: \: }$

$\sf{Therefore \: \: \: \: \: \: u = v}$

Differentiating both sides with respect v we get

$\displaystyle \sf{ \frac{du}{dv} = 1}$

RESULT

$\boxed{ \displaystyle \sf{ \: \: \frac{du}{dv} = 1 \: \: }}$

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