08 Chain Rule proof
by EngliSea on 2020-07-10
Definition of differentiation
f(g(x))▽x
= 「(f(g(x+h))−f(g(x)))/h :h⨠0」
= 「(f(g(x+h))−f(g(x)))/(g(x+h)−g(x))·(g(x+h)−g(x))/h :h⨠0」
= 「(f(g(x+h))−f(g(x)))/(g(x+h)−g(x)) :h⨠0」·「(g(x+h)−g(x))/h :h⨠0」


Let g(x+h)−g(x) = t, then 「t :h⨠0」 = 0 and g(x+h) = g(x) + t.
「(f(g(x+h))−f(g(x)))/(g(x+h)−g(x)) :h⨠0」
= 「(f(g(x)+t)−f(g(x)))/t :t⨠0」


= 「(f(g(x)+t)−f(g(x)))/t :t⨠0」·「(g(x+h)−g(x))/h :h⨠0」
= (f(g(x))▽g(x))·(g(x)▽x)