If y = sqrt{frac{1 + e^x}{1 - e^x}},then frac | vedclass

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(A) Given $y = \sqrt{\frac{1 + e^x}{1 - e^x}}$.Squaring both sides,we get $y^2 = \frac{1 + e^x}{1 - e^x}$.Differentiating both sides with respect to $

Vedclass · Accepted Answer

$\frac{e^x}{(1 - e^x)\sqrt{1 - e^{2x}}}$

Vedclass · Answer

(A) Given $y = \sqrt{\frac{1 + e^x}{1 - e^x}}$.Squaring both sides,we get $y^2 = \frac{1 + e^x}{1 - e^x}$.Differentiating both sides with respect to $x$ using the quotient rule:$2y \frac{dy}{dx} = \frac{(1 - e^x)(e^x) - (1 + e^x)(-e^x)}{(1 - e^x)^2}$$2y \frac{dy}{dx} = \frac{e^x - e^{2x} + e^x + e^{2x}}{(1 - e^x)^2} = \frac{2e^x}{(1 - e^x)^2}$$\frac{dy}{dx} = \frac{e^x}{y(1 - e^x)^2}$Substituting $y = \sqrt{\frac{1 + e^x}{1 - e^x}}$:$\frac{dy}{dx} = \frac{e^x}{\sqrt{\frac{1 + e^x}{1 - e^x}} (1 - e^x)^2} = \frac{e^x}{\sqrt{1 + e^x} \sqrt{1 - e^x} (1 - e^x)} = \frac{e^x}{(1 - e^x) \sqrt{(1 + e^x)(1 - e^x)}}$$\frac{dy}{dx} = \frac{e^x}{(1 - e^x) \sqrt{1 - e^{2x}}}$.

If $y = \sqrt{\frac{1 + e^x}{1 - e^x}}$,then $\frac{dy}{dx} = $

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