Intuition behind neglecting higher order differentials in visual proofs of the Product Rule

Question

In most visual explanations of the product rule (like this video), the justification for the remaining square ($df\cdot dg$) being neglected is that it is "very small". But the whole point of the intuition is that the $df \cdot g$ and $dg \cdot f$ are very small too.

As such, I'm interested in either an intuitive explanation of this, or an other intuitive explanation for the product rule that takes this into account.

Thanks in advance

Answer 1

It might help to compare the product rule against this (exact) equation for finite differences, where $\Delta f(x)$ means $f(x + \delta) - f(x)$:

$$ \Delta( fg(x) ) = g(x) \cdot \Delta f(x) + f(x+\delta) \cdot \Delta g(x)$$

If you find this equation helps you adequately intuit the product rule, then to answer the posted question observe the approximation

$$ g(x) \cdot \Delta f(x) + f(x+\delta) \cdot \Delta g(x) \approx g(x) \cdot \Delta f(x) + f(x) \cdot \Delta g(x) $$

ought to have negligible error; the difference between the two formulas is precisely $\Delta f(x) \Delta g(x)$.

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At some point you may be better off simply working through the algebra behind the proof of the product rule and basing your intuition off of the algebra directly, rather than looking for heuristic rationalizations.

Answer 2

Indeed $df\cdot g$ and $g\cdot df$ are very small but $df\cdot dg$ is even smaller.

For the intuition, let assume for examaple

• $\Delta f\approx \Delta g\approx10^{-3}$
• $f\approx g\approx10$

then

$$\Delta f \cdot g \approx10^{-2}\gg\Delta f \cdot \Delta g \approx10^{-6}$$

Answer 3

The derivative is about the first order term in the expanded functions, higher order terms vanish in the limit: \begin{align} (f(x) \, g(x))' &= \lim_{h\to 0} \frac{f(x+h)\, g(x+h) - f(x)\, g(x)}{h} \\ &= \lim_{h\to 0} \frac{(f(x) + f'(x) h + O(h^2))\, (g(x) + g'(x) h + O(h^2))) - f(x)\, g(x)}{h} \\ &= \lim_{h\to 0} \frac{f(x) \, g(x) + f'(x) \, g(x) h + f(x) \, g'(x) h + O(h^2) - f(x)\, g(x)}{h} \\ &= \lim_{h\to 0} f'(x) \, g(x) + f(x)\, g'(x) + O(h) \\ &= f'(x) \, g(x) + f(x) \, g'(x) \end{align} So too small in this context means being a term of order $k$ in $h$ with $k \ge 2$.