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Previous Year Question (PYQs)

The value of the derivative of the Sigmoid function given by

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

at $x=0$ is:





Solution

Given,

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

Differentiate with respect to $x$:

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

At $x=0$,

$e^{-2(0)}=e^0=1$

So,

$f'(0)=\dfrac{2}{(1+1)^2}$

$=\dfrac{2}{4}$

$=\dfrac{1}{2}$


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