Question

$y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right),\frac{dy}{dx}=$? at $x=0$?

Answer 1

$y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$
put $x=\sin \theta$ and Let $\cos \alpha =\left(\frac{5}{13}\right)\therefore \sin \alpha =\left(\frac{12}{13}\right)$
$\therefore y={\sin }^{-1}(\sin \theta .\cos \alpha +\cos \theta .\sin \alpha )={\sin }^{-1}\sin (\theta +\alpha )=\theta +\alpha ={\sin }^{-1}x+\alpha$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{(1-x^2)}}\therefore y^{\prime }\left(0\right)=1$