If y=sin-12x1+x2, t=sin-1 x1+x2, then dydx=

The given function is $y = \sin^{- 1} (\frac{2 x}{1 + x^{2}})$

Suppose that $x = \tan z$

$\begin{array}{rcl} y & = & \sin^{- 1} (\frac{2 \tan z}{1 + \tan^{2} z}) \\ = & \sin^{- 1} (\sin 2 z) \\ = & 2 z \\ = & 2 \tan^{- 1} x \end{array}$

Differenitate both sides

$\begin{array}{rcl} \frac{d y}{d x} & = & 2 \frac{d}{d x} (\tan^{- 1} x) \\ = & \frac{2}{1 + x^{2}} \end{array}$

Consider the function $t = \sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}})$

Suppose that $x = \tan z$

$\begin{array}{rcl} t & = & \sin^{- 1} (\frac{\tan z}{s e c z}) \\ = & \sin^{- 1} (\sin z) \\ = & z \\ = & \tan^{- 1} x \end{array}$

Differentiate both sides ieth respect to $x$

$\frac{d t}{d x} = \frac{1}{1 + x^{2}}$

Hence, $\frac{d y}{d x} = 2 \frac{d t}{d x}$

$If y = \sin^{- 1} (\frac{2 x}{1 + x^{2}}), t = \sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}}), then \frac{dy}{dx} =$