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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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