User:IssaRice/Proof that assumes the trick

Many proofs in mathematics depend on one or two "tricks". Some proofs (or ways of writing the proof) seem to deliberately hide or assume the trick so that the proof, while valid, is not very useful. (To the expert mathematician, the proof is obvious/they could have done it themselves, so what is the point of writing the proof? To the novice, the trick is assumed even if one might not know it; but the trick is what makes the proof work, so what is the point of seeing a proof that does not explain it?)

Consider Pugh's proof of the product rule for differentiation:

Since $\Delta (f\cdot g)=\Delta f\cdot g(x+\Delta x)+f(x)\cdot \Delta g$, continuity of $g$ at $x$ implies that

$${\frac {\Delta (f\cdot g)}{\Delta x}}={\frac {\Delta f}{\Delta x}}g(x+\Delta x)+f(x){\frac {\Delta g}{\Delta x}}\to f'(x)g(x)+f(x)g'(x)$$

What Pugh means is this:

On the one hand, $\Delta (f\cdot g):=(f\cdot g)(x+\Delta x)-(f\cdot g)(x)=f(x+\Delta x)g(x+\Delta x)-f(x)g(x)$.

We also have $\Delta f:=f(x+\Delta x)-f(x)$ and $\Delta g:=g(x+\Delta x)-g(x)$ so on the other hand we have

$${\begin{aligned}\Delta f\cdot g(x+\Delta x)+f(x)\cdot \Delta g&=(f(x+\Delta x)-f(x))\cdot g(x+\Delta x)+f(x)\cdot (g(x+\Delta x)-g(x))\\&=f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)g(x+\Delta x)-f(x)g(x)\\&=f(x+\Delta x)g(x+\Delta x)-f(x)g(x)\end{aligned}}$$

so the two are indeed equal.

What is the "trick" in this proof? Well, how would we have discovered that $\Delta (f\cdot g)=\Delta f\cdot g(x+\Delta x)+f(x)\cdot \Delta g$ if we didn't have Pugh to tell us? The trick is to add and subtract the same thing. We have

${\begin{aligned}f(x+\Delta x)g(x+\Delta x)-f(x)g(x)&=f(x+\Delta x)g(x+\Delta x)+\underbrace {(-f(x)g(x+\Delta x)+f(x)g(x+\Delta x))} _{=0}-f(x)g(x)\\&=g(x+\Delta x)(f(x+\Delta x)-f(x))+f(x)(g(x+\Delta x)-g(x))\end{aligned}}$