If $y = e^{\log \sin^{-1}x} + e^{\log \cos^{-1}x}$, 0 < x < 1, then

$\frac{dy}{dx} = 0$ $\frac{dy}{dx} = \frac{\pi}{2}$ $\frac{dy}{dx} = \frac{\pi}{3}$ does not exist

$\frac{dy}{dx} = 0$

$y = e^{\log \sin^{-1}x} + e^{\log \cos^{-1}x}$ $= \sin^{-1}x + \cos^{-1}x$ as $e^{\log a} = a$ as $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ $y = \frac{\pi}{2}$ so $\frac{dy}{dx} = 0$