frac{d}{dx} [operatorname{cosech}^{-1}(tan 2x | vedclass

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Question

(C) Let $y = \operatorname{cosech}^{-1}(	an 2x)$.Using the chain rule, $\frac{dy}{dx} = \frac{d}{dx} [\operatorname{cosech}^{-1}(	an 2x)]$.Recall th

Vedclass · Accepted Answer

$-2|\operatorname{cosec} 2x|$

Vedclass · Answer

(C) Let $y = \operatorname{cosech}^{-1}(	an 2x)$.Using the chain rule, $\frac{dy}{dx} = \frac{d}{dx} [\operatorname{cosech}^{-1}(	an 2x)]$.Recall the formula: $\frac{d}{dx} \operatorname{cosech}^{-1}(u) = \frac{-1}{|u| \sqrt{1+u^2}} \cdot \frac{du}{dx}$.Here, $u = 	an 2x$, so $\frac{du}{dx} = 2 \sec^2 2x$.Substituting these into the formula:$\frac{dy}{dx} = \frac{-1}{|	an 2x| \sqrt{1 + 	an^2 2x}} \cdot 2 \sec^2 2x$.Since $1 + 	an^2 2x = \sec^2 2x$, we have $\sqrt{1 + 	an^2 2x} = |\sec 2x|$.$\frac{dy}{dx} = \frac{-1}{|	an 2x| \cdot |\sec 2x|} \cdot 2 \sec^2 2x$.$\frac{dy}{dx} = \frac{-2 \sec^2 2x}{|\frac{\sin 2x}{\cos 2x}| \cdot |\frac{1}{\cos 2x}|} = \frac{-2 \sec^2 2x}{|\frac{\sin 2x}{\cos^2 2x}|} = \frac{-2 \sec^2 2x \cdot |\cos^2 2x|}{|\sin 2x|}$.Since $\sec^2 2x \cdot \cos^2 2x = 1$, this simplifies to:$\frac{dy}{dx} = \frac{-2}{|\sin 2x|} = -2 |\operatorname{cosec} 2x|$.

$\frac{d}{dx} [\operatorname{cosech}^{-1}(\tan 2x)] = $

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