Method

• 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*}