Find,$\cfrac{{dy}}{{dx}},ify = {\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt {1 - {x^2}} , - 1 \le x \le 1.$

Here,$y = {\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt {1 - {x^2}} = u + v$
Let $u = {\sin ^{ - 1}}x$ and $v = {\sin ^{ - 1}}\sqrt {1 - {x^2}}$

$\Rightarrow$ $\cfrac{{du}}{{dx}} = \cfrac{1}{{\sqrt {1 - {x^2}} }}$ and for $v = {\sin ^{ - 1}}\sqrt {1 - {x^2}}$

Putting $x = \cos \theta$, we get

$v = {\sin ^{ - 1}}\sqrt {1 - {{\cos }^2}\theta } = {\sin ^{ - 1}}\sqrt {{{\sin }^2}\theta }$
$= {\sin ^{ - 1}}(\sin \theta ) = \theta = {\cos ^{ - 1}}x$

therefore, $\cfrac{{dv}}{{dx}} = \cfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}$

So, $\cfrac{{dy}}{{dx}} = \cfrac{{du}}{{dx}} + \cfrac{{dv}}{{dx}} = \cfrac{1}{{\sqrt {1 - {x^2}} }} + \cfrac{{ - 1}}{{\sqrt {1 - {x^2}} }} = 0$