Why does the corner term vanish in the geometric derivation of the product rule?

Conditions

• Functions $f$ and $g$ are differentiable at the point.
• The input increment $h$ approaches zero.
• The geometric model assumes a rectangle with sides $f(x)$ and $g(x)$.

Reasoning, step by step

1. Visualize the area change as two strips and a small corner rectangle.
2. Express the exact increment as $g\Delta f + f\Delta g + \Delta f\Delta g$.
3. Note that differentiability implies $\Delta f \approx f'h$ and $\Delta g \approx g'h$.
4. Substitute these into the corner term: $\Delta f \Delta g \approx (f'h)(g'h) = f'g'h^2$.
5. Divide the total increment by $h$ to form the difference quotient.
6. Observe that the corner contribution is $f'g'h$, which vanishes as $h \to 0$.
7. Conclude that the derivative is the sum of the strip contributions: $f'g + fg'$.

Example

If $f(x)=x$ and $g(x)=x^2$ at $x=1$, a small $h$ makes $\Delta f \approx h$ and $\Delta g \approx 2h$. The corner area is roughly $2h^2$. Divided by $h$, it is $2h$, which goes to 0.

Common misconceptions

• Thinking the corner term is negligible because it is visually small, rather than because it is higher-order in $h$.
• Believing the product rule is simply the product of derivatives $f'g'$, which misses the strip terms entirely.
• Assuming the geometric argument fails for negative function values; the algebra holds for signed values even if the picture uses positive lengths.

Watch the explanation

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