GATE 2024 DA Q.33: solution and answer

GATE 2024 · DAQ.33 · 1 mark · NAT

Let $f : R \to R$ be the function $f (x) = \frac{1}{1 + e ^{- x}}$ . The value of the derivative of $f$ at $x$ where $f (x) = 0.4$ is ______ (rounded off to two decimal places). Note: $R$ denotes the set of real numbers.

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Answer and solution

ANSWER

The derivative of $f$ at the point where $f (x) = 0.4$ is $0.24$ .

Official answer: 0.24Source: Official answer key, GATE 2024 DA (IISc Bengaluru)

Recognise $f (x) = \frac{1}{1 + e ^{- x}}$ as the logistic sigmoid, defined for every real $x$ and taking values in $(0, 1)$ , so the value $f (x) = 0.4$ is attained at exactly one $x$ .
Differentiate with the chain rule. Write $f (x) = (1 + e^{- x})^{- 1}$ , so $f^{'} (x) = - (1 + e^{- x})^{- 2} \cdot (- e^{- x}) = \frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}$ .
Split the fraction: $f^{'} (x) = \frac{1}{1 + e ^{- x}} \cdot \frac{e ^{- x}}{1 + e ^{- x}}$ .
Since $\frac{e ^{- x}}{1 + e ^{- x}} = \frac{( 1 + e ^{- x} ) - 1}{1 + e ^{- x}} = 1 - \frac{1}{1 + e ^{- x}}$ , this gives $f^{'} (x) = f (x) (1 - f (x))$ .
The derivative depends only on the value $f (x)$ , so there is no need to solve for $x$ (it is $x = ln (2/3)$ , which is not required).
At $f (x) = 0.4$ : $f^{'} (x) = 0.4 \times 0.6 = 0.24$ .

KEY IDEA

For the logistic sigmoid, $f^{'} (x) = f (x) (1 - f (x))$ , so the derivative can be read straight from the function value without finding $x$ .

WATCH OUT FOR

Solving $\frac{1}{1 + e ^{- x}} = 0.4$ for $x$ and then substituting into the derivative is slower and invites rounding errors. Another slip is dropping the minus sign from the chain rule on $e^{- x}$ , or forgetting the square in the denominator.

From the GATE 2024 DA paper. Every question of this paper · Download the full paper · All 6 Limits & continuity questions, with notes

GATE 2024 DA Q.33: solution and answer

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