06 Quotient Rule proof
by EngliSea on 2020-07-09
(f(x)/g(x))▽x
= 「(f(x+h)/g(x+h)−f(x)/g(x))/h :h⨠0」
= 「(f(x+h)·g(x)−f(x)·g(x+h))/(g(x+h)·g(x))/h :h⨠0」 ➊
= 「(f(x+h)·g(x)−f(x)·g(x+h))/h/(g(x+h)·g(x)) :h⨠0」
= 「(f(x+h)·g(x)−f(x)·g(x+h))/h :h⨠0」
· 「1/(g(x+h)·g(x)) :h⨠0」
= 「((f(x+h)−f(x))·g(x)−f(x)·(g(x+h)−g(x)))/h :h⨠0」 ➋
· 「1/(g(x+h)·g(x)):h⨠0」
= 「((f(x+h)−f(x))/h·g(x)−f(x)·(g(x+h)−g(x))/h) :h⨠0」 ➌
· 「1/(g(x+h)·g(x)):h⨠0」
= (「(f(x+h)−f(x))/h:h⨠0」·g(x)−f(x)·「(g(x+h)−g(x))/h:h⨠0」)
· 「1/(g(x+h)·g(x)):h⨠0」
= ((f(x)▽x)·g(x)−f(x)·(g(x)▽x))/(g(x))²

➊ f(x+h)/g(x+h)−f(x)/g(x)
= (f(x+h)·g(x)−f(x)·g(x+h))/(g(x+h)·g(x))

➋ f(x+h)·g(x)−f(x)·g(x+h)
= f(x+h)·g(x)−f(x)·g(x)+f(x)·g(x)−f(x)·g(x+h)
= f(x+h)·g(x)−f(x)·g(x)−(f(x)·g(x+h)−f(x)·g(x))
= (f(x+h)−f(x))·g(x)−f(x)·(g(x+h)−g(x))

➌ ((f(x+h)−f(x))·g(x)−f(x)·(g(x+h)−g(x)))/h
= (f(x+h)−f(x))/h·g(x)−f(x)·(g(x+h)−g(x))/h