Sum and product
The familiar rules are not axioms — they are theorems, each provable from the difference quotient and the algebra of limits. The sum rule is immediate because limits add. The product rule needs one clever step: adding and subtracting the same term to split a stubborn quotient into two manageable pieces.
Product rule: if f, g are differentiable at a, then (fg)'(a) = f'(a)g(a) + f(a)g'(a). Difference quotient of fg at a (h ≠ 0): ( f(a+h)g(a+h) - f(a)g(a) ) / h Add and subtract f(a+h)g(a): = ( f(a+h)g(a+h) - f(a+h)g(a) + f(a+h)g(a) - f(a)g(a) ) / h = f(a+h) · ( g(a+h) - g(a) ) / h + ( f(a+h) - f(a) ) / h · g(a) Let h -> 0: f(a+h) -> f(a) (f is continuous at a, since differentiable) (g(a+h)-g(a))/h -> g'(a) (f(a+h)-f(a))/h -> f'(a) By algebra of limits: (fg)'(a) = f(a)·g'(a) + f'(a)·g(a). ∎
Quotient and chain
The quotient rule follows the same pattern, once you know 1/g is differentiable wherever g(a) ≠ 0 (continuity of g keeps g(a + h) away from 0 nearby). The chain rule is the deep one. The naive proof multiplies and divides by g(x) − g(a) — but that can be zero infinitely often even when g′(a) ≠ 0, so the naive step is not well-defined. The honest fix uses a correction-function argument.
Chain rule: with g differentiable at a and f differentiable at b = g(a),
(f ∘ g)'(a) = f'(b) · g'(a).
Key device: define a function E (the 'error') near b by
E(y) = ( f(y) - f(b) ) / ( y - b ) - f'(b) for y ≠ b,
E(b) = 0.
Then f(y) - f(b) = ( f'(b) + E(y) ) · ( y - b ) -- and this holds even at y = b
(both sides are 0). Because f is differentiable at b, E(y) -> 0 as y -> b, so E is
continuous at b.
Now put y = g(x), b = g(a):
f(g(x)) - f(g(a)) = ( f'(b) + E(g(x)) ) · ( g(x) - g(a) ).
Divide by x - a (x ≠ a):
( f(g(x)) - f(g(a)) ) / (x - a)
= ( f'(b) + E(g(x)) ) · ( g(x) - g(a) ) / (x - a).
Let x -> a: g continuous so g(x) -> b, hence E(g(x)) -> E(b) = 0;
(g(x)-g(a))/(x-a) -> g'(a).
Therefore the limit is ( f'(b) + 0 ) · g'(a) = f'(g(a)) · g'(a). ∎