The value of the derivative of Sigmoid function given by…
2019
The value of the derivative of Sigmoid function given by
\(f(x)=\dfrac{1}{1+e^{-2x}}\)
at \(𝑥=0\) is
Answer: B. \(1 \over 2\) — Solution: Compute the derivative and evaluate at x = 0. Let f(x) = 1/(1+e^{-2x}). Use the chain rule. Set u = e^{-2x}, so f = 1/(1+u). Then df/du = -1/(1+u)^2…
- A.
\(0\) - B.
\(1 \over 2\) - C.
\(1 \over 4\) - D.
\(\infty\)
Attempted by 11 students.
Show answer & explanation
Correct answer: B
Solution: Compute the derivative and evaluate at x = 0.
Let f(x) = 1/(1+e^{-2x}). Use the chain rule.
Set u = e^{-2x}, so f = 1/(1+u). Then df/du = -1/(1+u)^2 and du/dx = -2 e^{-2x}.
Therefore df/dx = (df/du)(du/dx) = (-1/(1+u)^2)(-2 e^{-2x}) = 2 e^{-2x}/(1+e^{-2x})^2.
Evaluate at x = 0: e^{-2·0} = 1, so f'(0) = 2·1/(1+1)^2 = 2/4 = 1/2.
Alternate compact form: f'(x) = 2 f(x)(1 - f(x)). Since f(0) = 1/2, this gives f'(0) = 2*(1/2)*(1/2) = 1/2.
Final answer: The derivative at x = 0 is 1/2.
A video solution is available for this question — log in and enroll to watch it.