If y = log left( frac{1 + sqrt{x}}{1 - sqrt{x | vedclass

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(B) Given $y = \log \left( \frac{1 + \sqrt{x}}{1 - \sqrt{x}} \right)$.Using logarithmic properties,$y = \log(1 + \sqrt{x}) - \log(1 - \sqrt{x})$.Diffe

Vedclass · Accepted Answer

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

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(B) Given $y = \log \left( \frac{1 + \sqrt{x}}{1 - \sqrt{x}} \right)$.Using logarithmic properties,$y = \log(1 + \sqrt{x}) - \log(1 - \sqrt{x})$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [\log(1 + \sqrt{x})] - \frac{d}{dx} [\log(1 - \sqrt{x})]$$\frac{dy}{dx} = \frac{1}{1 + \sqrt{x}} \cdot \frac{1}{2\sqrt{x}} - \frac{1}{1 - \sqrt{x}} \cdot \left( -\frac{1}{2\sqrt{x}} \right)$$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left( \frac{1}{1 + \sqrt{x}} + \frac{1}{1 - \sqrt{x}} \right)$$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left( \frac{1 - \sqrt{x} + 1 + \sqrt{x}}{(1 + \sqrt{x})(1 - \sqrt{x})} \right)$$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left( \frac{2}{1 - x} \right)$$\frac{dy}{dx} = \frac{1}{\sqrt{x}(1 - x)}$.

If $y = \log \left( \frac{1 + \sqrt{x}}{1 - \sqrt{x}} \right)$,then $\frac{dy}{dx} = $

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