$\dfrac{dy}{dx}$ for $y=\sin^{-1}(\cos x)$, where $x\in(0,\pi)$, is $-1$
$\dfrac{dy}{dx}$ for $y=\sin^{-1}(\cos x)$, where $x\in(0,2\pi)$, is $1$
$\dfrac{dy}{dx}$ for $y=\cos^{-1}(\sin x)$, where $x\in(-\pi,\pi/2)$, is $-1$
$\dfrac{dy}{dx}$ for $y=\cos^{-1}(\sin x)$, where $x\in(-\pi/2,3\pi/2)$, is $-1$
Step-by-Step Solution
Key Concept: $\sin^{-1}(\cos x)=\pi/2-x$ for $x\in(0,\pi)$; derivative $=-1$
$y=\sin^{-1}(\cos x)=\pi/2-x$ for $x\in(0,\pi)$. $dy/dx=-1$. Option 1 is true.
Correct Answer: 1