frac{d}{d x}left(operatorname{cosec}^{-1} e^x | vedclass

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Question

(C) Let $y = \operatorname{cosec}^{-1}(e^x)$.Using the chain rule,we differentiate with respect to $x$:$\frac{d y}{d x} = \frac{d}{d x}(\operatorname{

Vedclass · Accepted Answer

$\frac{-1}{e^x \sqrt{e^{2 x}-1}}$

Vedclass · Answer

(C) Let $y = \operatorname{cosec}^{-1}(e^x)$.Using the chain rule,we differentiate with respect to $x$:$\frac{d y}{d x} = \frac{d}{d x}(\operatorname{cosec}^{-1}(e^x))$We know that $\frac{d}{d u}(\operatorname{cosec}^{-1} u) = \frac{-1}{|u| \sqrt{u^2 - 1}}$.Here,$u = e^x$,so $\frac{d u}{d x} = e^x$.Applying the chain rule:$\frac{d y}{d x} = \frac{-1}{|e^x| \sqrt{(e^x)^2 - 1}} \cdot \frac{d}{d x}(e^x)$Since $e^x > 0$ for all real $x$,$|e^x| = e^x$.$\frac{d y}{d x} = \frac{-1}{e^x \sqrt{e^{2 x} - 1}} \cdot e^x$$\frac{d y}{d x} = \frac{-1}{\sqrt{e^{2 x} - 1}}$Therefore,the correct option is $C$.

$\frac{d}{d x}\left(\operatorname{cosec}^{-1} e^x\right) = $ . . . . . .

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