If y = operatorname{cosec}^{-1}(x) and frac{d | vedclass

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(C) The derivative of $y = \operatorname{cosec}^{-1}(x)$ is given by $\frac{dy}{dx} = \frac{-1}{|x| \sqrt{x^2-1}}$.For the derivative to be defined,th

Vedclass · Accepted Answer

$y \in \left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right)$

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(C) The derivative of $y = \operatorname{cosec}^{-1}(x)$ is given by $\frac{dy}{dx} = \frac{-1}{|x| \sqrt{x^2-1}}$.For the derivative to be defined,the expression inside the square root must be strictly positive,i.e.,$x^2 - 1 > 0$,which implies $|x| > 1$.However,the range of $y = \operatorname{cosec}^{-1}(x)$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}$.Since the derivative $\frac{dy}{dx}$ is defined for all $x$ in the domain of the function excluding the endpoints where the derivative might not exist or is undefined,we consider the open interval for $y$.The values of $y$ corresponding to the domain of the derivative are $y \in \left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right)$.

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

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