यदि y = operatorname{Tanh}^{-1} sqrt{frac{1-x | vedclass

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(B) दिया गया है $y = \operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}$.माना $x = \cos 	heta$,तब $	heta = \cos^{-1} x$.तब $\sqrt{\frac{1-x}{1+x}} = \s

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

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(B) दिया गया है $y = \operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}$.माना $x = \cos 	heta$,तब $	heta = \cos^{-1} x$.तब $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = \sqrt{\frac{2 \sin^2 (	heta/2)}{2 \cos^2 (	heta/2)}} = 	an(	heta/2)$.अतः,$y = \operatorname{Tanh}^{-1} (	an(	heta/2))$.सर्वसमिका $\operatorname{Tanh}^{-1} z = \frac{1}{2} \ln \left( \frac{1+z}{1-z} ight)$ का उपयोग करने पर:$y = \frac{1}{2} \ln \left( \frac{1+	an(	heta/2)}{1-	an(	heta/2)} ight) = \frac{1}{2} \ln \left( 	an \left( \frac{\pi}{4} + \frac{	heta}{2} ight) ight) = \frac{1}{2} \left( \ln 	an \left( \frac{\pi}{4} + \frac{\cos^{-1} x}{2} ight) ight)$.$x$ के सापेक्ष अवकलन करने पर:$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{	an(\frac{\pi}{4} + \frac{	heta}{2})} \cdot \sec^2(\frac{\pi}{4} + \frac{	heta}{2}) \cdot \frac{1}{2} \cdot \frac{d}{dx}(\cos^{-1} x)$.$\frac{dy}{dx} = \frac{1}{4} \cdot \frac{1}{\sin(\frac{\pi}{4} + \frac{	heta}{2}) \cos(\frac{\pi}{4} + \frac{	heta}{2})} \cdot \left( -\frac{1}{\sqrt{1-x^2}} ight)$.$\frac{dy}{dx} = \frac{1}{2 \sin(\frac{\pi}{2} + 	heta)} \cdot \left( -\frac{1}{\sqrt{1-x^2}} ight) = \frac{1}{2 \cos 	heta} \cdot \left( -\frac{1}{\sqrt{1-x^2}} ight) = -\frac{1}{2x \sqrt{1-x^2}}$.

यदि $y = \operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}$ है,तो $\frac{dy}{dx} = $

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