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Find \( \frac {dy}{dx} \) in the following: \( y = sin^{-1} \left(\frac{2x}{1 + x^2}\right) \)

$\begin{array}{1 1} \large \frac{2}{1+x^2} \\ \large \frac{2x}{1+x^2}\\ -\large \frac{2}{1+x^2} \\ -\large \frac{2x}{1+x^2} \end{array} $

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  • $\; \sin 2x = \large \frac{2\tan x}{1 + \tan^2x}$
  • $\; \large \frac{d(tan^{-1}x)}{dx} $$= \large\frac{1}{1+x^2}$
Given $y = sin^{-1} \left(\large\frac{2x}{1 + x^2}\right)$.
Let us substitute $\theta = \tan^{-1} x \rightarrow y = sin^{-1} \left(\large\frac{2\tan \theta}{1 + \tan^2 \theta}\right)$.
We know that $\; \sin 2x = \large \frac{2\tan x}{1 + \tan^2x}$
Therefore, $y = \sin^{-1} \sin 2\theta = 2\theta = 2\tan^{-1} x$
Also, we know that $\; \large \frac{d(tan^{-1}x)}{dx} $$= \large\frac{1}{1+x^2}$
Differentating, we get:
$\Rightarrow dy = 2dx.\large \frac{1}{1+x^2}$
$\Rightarrow \large \frac{dy}{dx} = \frac{2}{1+x^2}$
answered Apr 5, 2013 by balaji.thirumalai
 

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