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把求导法则做严格

我们从定义出发证明加法、乘积、商与链式法则——并找出链式法则中唯一真正微妙的地方。

加法与乘积

那些熟悉的法则不是公理——它们是定理,每一条都能从差商极限的代数运算证出。加法法则是显然的,因为极限可加。乘积法则需要一个巧妙的步骤:加上又减去同一项,把一个棘手的商拆成两个好处理的部分。

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).   ∎
加减同一项的技巧,再用 f 的连续性处理 f(a+h) 这个因子。

商与链式

商法则遵循同样的套路,只要你知道在 g(a) ≠ 0 处 1/g 可微(g 的连续性使得附近的 g(a + h) 不为 0)。链式法则才是深刻的那个。朴素的证明会乘以又除以 g(x) − g(a)——但即便 g′(a) ≠ 0,这个量也可能无穷多次为零,于是朴素的步骤并非良定义的。诚实的补救用的是一个修正函数的论证。

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).  ∎
误差函数 E 吸收了 y − b = 0 的麻烦:恒等式对所有 y 都成立,无需做除法。