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Memory Based GATE DA 2024 | Question: 7

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$f'(x) = \dfrac{e^{-x}}{(1+e^{-x})^2}$ $----(1)$

We have to find the derivative value at $x$ such that $f(x)= 0.4$

$\dfrac{1}{1+e^{-x}} = 0.4$

$1+ e^{-x} = \dfrac{1}{0.4} = 2.5$

$e^{-x} = 2.5 - 1 = 1.5 $

Substitute $e^{-x}$ value in $(1)$ then it will be,

$f'(x) = \dfrac{1.5}{(1+1.5)^2}$

            $=0.24$

So the answer is $0.24$.
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