Explanation

(b)
Given, $y = \sin^{- 1}\left( \frac{5x + 12\sqrt{1 - x^{2}}}{13} \right)$
Put $\ x = \sin\theta$
$\therefore\ \ \ \ y = \sin^{- 1}\left( \frac{5\sin\theta + 12\cos\theta}{13} \right)$
=${\ sin}^{- 1}{\{\sin{(\theta + \alpha)\}\left( Put\cos{\alpha\ = \frac{5}{13}} \right)}}$
$\Rightarrow \ \ y =$ $\theta + \alpha\ = \sin^{- 1}{x + \alpha}$
On differentiating w.r.t. $x,$ we get
$\ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^{2}}}$