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The Jacobian

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A gradient is for a function with one output. When a function has several outputs, each one has its own gradient, and stacking them as rows gives the Jacobian:

Row $i$ is the gradient of output $i$ . So a function from to has an $(m, n)$ Jacobian — outputs down, inputs across. Getting that orientation wrong is the classic transpose bug.

Every layer of a neural network is a vector function, and the chain rule for them is Jacobian multiplication: $J_{total} = J_{3} J_{2} J_{1}$ . Backpropagation never forms these matrices — for a layer with a thousand units the Jacobian would have a million entries — but it computes exactly the vector-Jacobian products that this expression implies. Knowing that is what makes dX = dZ @ W.T stop looking arbitrary.

This problem uses

f (x, y) = [x^{2} y 5 x + sin y], J = [2 x y 5 x^{2} cos y]

at (x, y) = (1, 0):
  J = [[2*1*0, 1^2 ],     =  [[0, 1],
       [5,     cos 0]]        [5, 1]]

Your task:

Implement jacobian_at(x, y) returning the $2 \times 2$ Jacobian as a nested list [[df1/dx, df1/dy], [df2/dx, df2/dy]].

Example Tests

At (1, 0) the sine row contributes cos(0) = 1

Input: {"x":1,"y":0}

Expected: [[0,1],[5,1]]

At (2, 1) the top row is [2xy, x^2] = [4, 4]

Input: {"x":2,"y":1}

Expected: [[4,4],[5,0.5403]]

The constant 5 appears in every Jacobian regardless of position

Input: {"x":0,"y":0}

Expected: [[0,0],[5,1]]

Python

import numpy as np


def jacobian_at(x, y):
    """
    Jacobian of f(x, y) = [x^2*y, 5x + sin(y)].

    Args:
        x: first coordinate
        y: second coordinate

    Returns:
        2x2 nested list: rows are outputs, columns are inputs
    """
    # YOUR CODE HERE
    pass

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