Why does the higher-order corner term $\Delta f \Delta g$ vanish in the limit when deriving the product rule?

Conditions

• Functions $f$ and $g$ are differentiable at the point of interest
• Input increment $h$ approaches zero

Reasoning, step by step

1. Identify the exact area increment: $\Delta(fg) = g\Delta f + f\Delta g + \Delta f \Delta g$.
2. Form the difference quotient by dividing by $h$: $\frac{\Delta(fg)}{h} = g\frac{\Delta f}{h} + f\frac{\Delta g}{h} + \frac{\Delta f \Delta g}{h}$.
3. Apply the definition of differentiability: $\Delta f \approx f' h$ and $\Delta g \approx g' h$ for small $h$.
4. Substitute these approximations into the corner term: $\frac{(f'h)(g'h)}{h} = f'g'h$.
5. Take the limit as $h \to 0$: Since $f'g'$ is finite, $f'g'h \to 0$.
6. Conclude that only the first two terms remain, yielding $(fg)' = f'g + fg'$.

Example

Script segment at 350s states: 'Differentiability makes both Δf and Δg of order h, so their corner product divided by h tends to zero. The surviving terms give $f^{\prime} g+fg^{\prime}$.'

Common misconceptions

• Believing the corner term represents a significant error rather than a negligible higher-order infinitesimal.
• Assuming the term disappears simply because it is multiplied together, ignoring the division by $h$ required for the derivative definition.

Watch the explanation

Explore next

• How does this geometric model extend to the product of three functions?
• What happens if one function is not differentiable?