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

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

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} $$= \large\frac{1}{1+x^2} Given y = cos^{-1} \left(\large \frac{1 - x^2}{1 + x^2}\right),$$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 = cos^{-1} \left(\large \frac{1 - \tan^2\theta}{1 + \tan^2\theta}\right), $$0 < x < 1 \cos 2x = \large\frac{1 - tan^2x}{1 + tan^2x} \Rightarrow y = cos^{-1} cos 2\theta = 2\theta = 2 \tan^{-1} x Differentiating both sides: \Rightarrow dy = 2 \large\frac{1}{1+x^2}$$\;dx$
Remember $\; \large \frac{d(tan^{-1}x)}{dx}$$= \large\frac{1}{1+x^2}$
$\Rightarrow \large\frac{dy}{dx} = \frac{2}{1+x^2}$