frac{d}{dx} left{ log left( frac{e^x}{1 + e^x | vedclass

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(D) Let $y = \log \left( \frac{e^x}{1 + e^x} \right)$.Using the property of logarithms,$\log \left( \frac{a}{b} \right) = \log a - \log b$,we get:$y =

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $y = \log \left( \frac{e^x}{1 + e^x} \right)$.Using the property of logarithms,$\log \left( \frac{a}{b} \right) = \log a - \log b$,we get:$y = \log(e^x) - \log(1 + e^x) = x - \log(1 + e^x)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(\log(1 + e^x))$.$\frac{dy}{dx} = 1 - \frac{1}{1 + e^x} \cdot \frac{d}{dx}(1 + e^x)$.$\frac{dy}{dx} = 1 - \frac{e^x}{1 + e^x}$.$\frac{dy}{dx} = \frac{1 + e^x - e^x}{1 + e^x} = \frac{1}{1 + e^x}$.Since $\frac{1}{1 + e^x}$ is not among the given options,the correct answer is $(D)$.

$\frac{d}{dx} \left\{ \log \left( \frac{e^x}{1 + e^x} \right) \right\} = $

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