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If $y = \tan^{-1}\left(\frac{2x}{1-x^{2}} \right )$ then $\frac{\mathrm{d} y}{\mathrm{d} x} =$ ? $(0<x <1)$
Option: 1
$\frac{1}{1 + x^{2}}$

Option: 2
$-\frac{1}{1 + x^{2}}$

Option: 3
$\frac{2}{1 + x^{2}}$

Option: 4
$-\frac{2}{1 + x^{2}}$

Answers (1)

best_answer

As we have learned

Differentiation of inverse trigonometric function -

Let

$y=sin^{-1}(\frac{2x}{1+x^{3}})$

$put\:\:x=tan\theta$

$\therefore \:\:sin^{-1}(\frac{2tan\theta}{1+tan^{2}\theta})$

$=sin^{-1}sin(2\theta)$

$=2\theta=2tan^{-1}x$

$\therefore \:\:\frac{dy}{dx}=\frac{2}{1+x^{2}}$

- wherein

Use : for

$\frac{2x}{1+x^{2}},\:\:\frac{1-x^{2}}{1+x^{2}}\:\:and\:\:\frac{2x}{1-x^{2}}$

$put\:\:x=tan\theta\:\:and\:gives:\:sin2\theta$

$cos2\theta\:\:and\;\;tan2\theta\:\:respectively$

Let x = $\tan \theta \Rightarrow y = \tan ^{-1}\left ( \frac{2\tan \theta }{1-\tan ^2 \theta} \right )= \tan ^{-1}(\tan 2\theta )= 2 \theta$

$\Rightarrow y = 2\tan ^{-1} x \Rightarrow \frac{dy}{dx}= \frac{2}{1+x^2}$

Posted by

rishi.raj

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