• # question_answer If $y=si{{n}^{-1}}\left( \frac{5x+12\sqrt{1-{{x}^{2}}}}{13} \right)$ then $\frac{dy}{dx}$is equa.to A)  $\frac{1}{\sqrt{1-{{x}^{2}}}}$                       B)  $\frac{-1}{\sqrt{1-{{x}^{2}}}}$ C)  $\frac{2}{\sqrt{1-{{x}^{2}}}}$                       D)  $\frac{-2}{\sqrt{1-{{x}^{2}}}}$

[b] putting $x=\cos \theta$ $\therefore y=soi{{n}^{-1}}\left\{ \frac{5.\cos \theta +12.\sin \theta }{13} \right\}={{\sin }^{-1}}\left[ \left( \frac{5}{13} \right).\cos \theta +\frac{12}{13} \right].\sin \theta$Let $\frac{5}{13}=\sin \alpha ,$ then$\Rightarrow \frac{12}{13}=\cos \alpha ,$ $\therefore y={{\sin }^{-1}}\{\sin (\alpha +\theta )\}=\alpha +\theta$ Differentiating w.r.t. x, we have $\frac{dy}{d\theta }=\frac{d\theta }{dx}=\frac{d}{dx}({{\cos }^{-1}}x)=\frac{-1}{\sqrt{1-{{x}^{2}}}}$ Hence, option [b] is correct.