If y = sinh^{-1}left(frac{1-x}{1+x}right),the | vedclass

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(A) Given $y = \sinh^{-1}\left(\frac{1-x}{1+x}\right)$.Using the chain rule,$\frac{dy}{dx} = \frac{d}{du}(\sinh^{-1}(u)) \cdot \frac{du}{dx}$,where $u

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

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(A) Given $y = \sinh^{-1}\left(\frac{1-x}{1+x}\right)$.Using the chain rule,$\frac{dy}{dx} = \frac{d}{du}(\sinh^{-1}(u)) \cdot \frac{du}{dx}$,where $u = \frac{1-x}{1+x}$.The derivative of $\sinh^{-1}(u)$ is $\frac{1}{\sqrt{u^2+1}}$.Now,$\frac{du}{dx} = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$.Substituting these into the chain rule:$\frac{dy}{dx} = \frac{1}{\sqrt{(\frac{1-x}{1+x})^2 + 1}} \cdot \frac{-2}{(1+x)^2}$.$\frac{dy}{dx} = \frac{1}{\sqrt{\frac{(1-x)^2 + (1+x)^2}{(1+x)^2}}} \cdot \frac{-2}{(1+x)^2}$.$\frac{dy}{dx} = \frac{|1+x|}{\sqrt{1-2x+x^2+1+2x+x^2}} \cdot \frac{-2}{(1+x)^2}$.$\frac{dy}{dx} = \frac{|1+x|}{\sqrt{2+2x^2}} \cdot \frac{-2}{(1+x)^2} = \frac{|1+x|}{\sqrt{2}\sqrt{1+x^2}} \cdot \frac{-2}{(1+x)^2}$.Since $\frac{|1+x|}{(1+x)^2} = \frac{1}{|1+x|}$,we get:$\frac{dy}{dx} = \frac{-\sqrt{2}}{|1+x|\sqrt{1+x^2}}$.

If $y = \sinh^{-1}\left(\frac{1-x}{1+x}\right)$,then find $\frac{dy}{dx}$.

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