If $y=\operatorname{cosec}^{-1}(x)$ and $\frac{d y}{d x}=\frac{-1}{|x| \sqrt{x^2-1}}$, then

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Aicharya · Accepted Answer

It is given that, $y=\operatorname{cosec}^{-1} x$ $\text {and } \frac{d y}{d x}=\frac{-1}{|x| \sqrt{x^2-1}}$ $\because$ Domain of $\operatorname{cosec}^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$ and the $\frac{d y}{d x}$ is define for $x \in\left(-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right)$. Because for derivatives we should exclude the end points of a internal. Hence, option (c) is correct.

If \(y=\operatorname{cosec}^{-1}(x)\) and \(\frac{d y}{d x}=\frac{-1}{|x| \sqrt{x^2-1}}\), then

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