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

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

Toolbox:
• $\; \sin 2x = \large \frac{2\tan x}{1 + \tan^2x}$
• $\; \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