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Find \( \frac {dy}{dx} \) in the following:$ y = sin^{-1} \left(\frac{1 - x^2}{1 + x^2}\right), 0 < x < 1 $

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

1 Answer

Toolbox:
  • For inverse trigonometric functions, we first make a substitution to make it a standard trigonometric expression and then simplify the expression and differentiate it.
  • $\cos 2x = \large\frac{1 - tan^2x}{1 + tan^2x}$
  • $\large \frac{d(tan^{-1}x)}{dx} $$= \large\frac{1}{1+x^2}$
  • $cos 2\theta = sin (\frac{\pi}{2} -2\theta)$
Given $y = sin^{-1} \left(\large\frac{1 - x^2}{1 + x^2}\right)$$, 0 < x < 1$
For inverse trigonometric functions, we first make a substitution to make it a standard trigonometric expression and then simplify the expression and differentiate it.
Since we know that $\cos 2x = \large\frac{1 - tan^2x}{1 + tan^2x}$, we'll substitute such that the given equation can be simplified
Let $x = \tan \theta \rightarrow y = sin^{-1} \left(\large \frac{1 - \tan^2\theta}{1 + \tan^2\theta}\right), $$0 < x < 1$
$\cos 2x = \large\frac{1 - tan^2x}{1 + tan^2x}$
$\Rightarrow y = sin^{-1} cos 2\theta = sin^{-1}sin (\frac{\pi}{2} -2\theta) = \frac{\pi}{2} -2\theta$
$\Rightarrow y = \frac{\pi}{2} -2\tan^{-1}x$
Differentiating both sides:
$\Rightarrow dy = 0 - 2 \large\frac{1}{1+x^2}$$\;dx$
$\Rightarrow \large\frac{dy}{dx} = \frac{-2}{1+x^2}$
answered Apr 5, 2013 by balaji.thirumalai
 

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