GATE 2025 DA Q.42: solution and answer

GATE 2025 · DAQ.42 · 2 marks · MCQ

Consider the neural network shown in the figure with

inputs: $u, v$
weights: $a, b, c, d, e, f$
output: $y$

$R$ denotes the ReLU function, $R (x) = max (0, x)$ .

Given $u = 2$ , $v = 3$ ,
$a = 1, b = 1, c = 1, d = - 1, e = 4, f = - 1$ ,

which one of the following is correct?

GATE 2025 DA question 42: Feed-forward network. — Figure from the GATE 2025 DA paper. It is part of the question.

(A)
∂y/∂a = 8, ∂y/∂f = 0
(B)
∂y/∂a = 1, ∂y/∂f = 0
(C)
∂y/∂a = 1, ∂y/∂f = −1
(D)
∂y/∂a = 2, ∂y/∂f = −1

Answer and solution

ANSWER

Option (A): $\partial y / \partial a = 8$ and $\partial y / \partial f = 0$ .

Official answer: ASource: Official answer key, GATE 2025 DA (IIT Roorkee)

Read the figure: $a$ joins $u$ to the top ReLU unit, $b$ joins $v$ to the top unit, $c$ joins $u$ to the bottom unit, and $d$ joins $v$ to the bottom unit. The top and bottom units feed the output ReLU through $e$ and $f$ respectively. So $z_{1} = a u + b v$ , $h_{1} = R (z_{1})$ , $z_{2} = c u + d v$ , $h_{2} = R (z_{2})$ , $z_{3} = e h_{1} + f h_{2}$ and $y = R (z_{3})$ .
Top hidden unit: $z_{1} = 1 \cdot 2 + 1 \cdot 3 = 5 > 0$ , so $h_{1} = R (5) = 5$ .
Bottom hidden unit: $z_{2} = 1 \cdot 2 + (- 1) \cdot 3 = - 1 < 0$ , so $h_{2} = R (- 1) = 0$ .
Output unit: $z_{3} = 4 \cdot 5 + (- 1) \cdot 0 = 20 > 0$ , so $y = 20$ . The output ReLU is active, with $R^{'} (z_{3}) = 1$ .
The derivative of ReLU is $R^{'} (x) = 1$ for $x > 0$ and $0$ for $x < 0$ . The top unit has $z_{1} = 5 > 0$ , so $R^{'} (z_{1}) = 1$ .
Chain rule for $a$ : $\frac{\partial y}{\partial a} = R^{'} (z_{3}) \cdot e \cdot R^{'} (z_{1}) \cdot u = 1 \cdot 4 \cdot 1 \cdot 2 = 8.$
Chain rule for $f$ : $\frac{\partial y}{\partial f} = R^{'} (z_{3}) \cdot h_{2} = 1 \cdot 0 = 0.$ The bottom unit is inactive, so $h_{2} = 0$ and $f$ has no effect on $y$ at this input.
So $\partial y / \partial a = 8$ and $\partial y / \partial f = 0$ , which is option (A).

KEY IDEA

Run the forward pass to see which ReLUs are active. Then apply the chain rule through the active path only. A ReLU with a negative pre-activation has slope 0 and outputs 0, so it blocks the gradient and contributes nothing to the next layer.

WATCH OUT FOR

Forgetting the ReLU slopes and taking $\partial y / \partial f = h_{2}$ or $- 1$ (the weight's value) instead of the bottom unit's output, which is $0$ . Another slip is to give $\partial y / \partial a = u = 2$ or $1$ by leaving out the factor $e = 4$ from the downstream weight.

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 5 Neural networks questions, with notes

GATE 2025 DA Q.42: solution and answer

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