If f(x)=cosh ^{-1}left(frac{1-x}{1+x}right) i | vedclass

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(A) Given $f(x) = \cosh^{-1}\left(\frac{1-x}{1+x}\right)$.Using the derivative formula $\frac{d}{dx}(\cosh^{-1}(u)) = \frac{1}{\sqrt{u^2-1}} \cdot \fr

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$\frac{-1}{(1+x) \sqrt{-x}}$

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(A) Given $f(x) = \cosh^{-1}\left(\frac{1-x}{1+x}\right)$.Using the derivative formula $\frac{d}{dx}(\cosh^{-1}(u)) = \frac{1}{\sqrt{u^2-1}} \cdot \frac{du}{dx}$:$f'(x) = \frac{1}{\sqrt{\left(\frac{1-x}{1+x}\right)^2 - 1}} \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$$f'(x) = \frac{1}{\sqrt{\frac{(1-x)^2 - (1+x)^2}{(1+x)^2}}} \cdot \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2}$$f'(x) = \frac{1+x}{\sqrt{1 - 2x + x^2 - (1 + 2x + x^2)}} \cdot \frac{-1 - x - 1 + x}{(1+x)^2}$$f'(x) = \frac{1+x}{\sqrt{-4x}} \cdot \frac{-2}{(1+x)^2}$$f'(x) = \frac{1+x}{2\sqrt{-x}} \cdot \frac{-2}{(1+x)^2} = \frac{-1}{(1+x)\sqrt{-x}}$.

If $f(x)=\cosh ^{-1}\left(\frac{1-x}{1+x}\right)$ is well defined, then $f^{\prime}(x)=$

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