---

Method

Differentiate

\[\begin{equation*}
  f(x, y) = (x + y) (y + 3)
\end{equation*}
\]

at

\[\begin{align*}
  x & = 1 \\
  y & = 2
\end{align*}
\]

with respect to \(x\) and \(y\).

• Diagram
• Forward pass

  Compute forward

  \[\begin{align*}
    a & = x + y = 1 + 2 = 3 \\
    b & = y + 3 = 2 + 3 = 5 \\
    c & = ab = 3 \cdot 5 = 15
  \end{align*}
  \]

  getting

  \[\begin{equation*}
    f(x) = c = 15.
  \end{equation*}
  \]

• Backward pass

  Using cached \(a\), \(b\), and \(c\), we proceed backward:

  • Compute the gradients of \(c\) with respect to \(a\) and \(b\):

    \[\begin{align*}
      \pdv{c}{a} & = \pdv{a} ab = \pdv{a} 5a \eval{}_{b = 5} = 5 \\[1ex]
      \pdv{c}{b} & = \pdv{b} ab = \pdv{b} 3b \eval{}_{a = 3} = 3
    \end{align*}
    \]

  • Compute the gradients of \(a\) with respect to \(x\) and \(y\):

    \[\begin{align*}
      \pdv{a}{x} & = \pdv{x} (x + y) = \pdv{x} (x + 2) \eval{}_{y = 2} = 1 \\[1ex]
      \pdv{a}{y} & = \pdv{y} (x + y) = \pdv{x} (1 + y) \eval{}_{x = 1} = 1
    \end{align*}
    \]

  • Compute the gradients of \(b\) with respect to \(x\) and \(y\):

    \[\begin{align*}
      \pdv{b}{x} & = \pdv{x} (y + 3) = \pdv{x} (2 + 3) \eval{}_{y = 2} = 0 \\[1ex]
      \pdv{b}{y} & = \pdv{y} (y + 3) = \pdv{x} (y + 3) \eval{}_{x = 1} = 1
    \end{align*}
    \]

  • Use the chain rule to compute the gradients of \(c\) with respect to \(x\) and \(y\):

    \[\begin{align*}
      \pdv{c}{x} & = \pdv{c}{a} \pdv{a}{x} + \pdv{c}{b} \pdv{b}{x} = 5 \cdot 1 + 3 \cdot 0 = 5 \\[1ex]
      \pdv{c}{y} & = \pdv{c}{a} \pdv{a}{y} + \pdv{c}{b} \pdv{b}{y} = 5 \cdot 1 + 3 \cdot 1 = 8
    \end{align*}
    \]