Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was reading Stewart's Calculus text and came across the Implicit Function Theorem for the Chain Rule with more than two variables:

$$\frac{dy}{dx} = - \dfrac {{\dfrac{\partial F}{\partial x}}} {\dfrac{\partial F}{\partial y}} = -\frac{F_x}{F_y} \tag{1}$$

He states that

To derive (1), we assumed $F(x,y)=0$ defines $y$ implicitly as a function of $x$. The IFT, proved in advanced calculus, gives conditions under which this assumption is valid: It states that if $F$ is defined on a disk containing $(a,b)$, where $F(a,b) = 0, \ F_y(a,b) \ne 0,$ and $F_x$ and $F_y$ are continuous on the disk, then the equation $F(x,y) = 0$ defines $y$ as a function of $x$ near the point $(a,b)$ and the derivative of this function is given by (1).

While he says it is proved in advanced calculus, is there a simpler proof? I am looking for proofs (perhaps one that only uses multivariable calculus knowledge) for this and I am also wondering what are the uses of this theorem (IFT) in higher branches of mathematics as well as other disciplines (engineering, economics, etc.).

share|cite|improve this question
A google search gave… – anonymous Jan 11 '13 at 4:00
Thanks for the MO link - reading now. – Joe Jan 11 '13 at 4:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.