"Let $\mathbb{F}$ be a real-valued continuously differentiable function defeined in a neighborhood of $(X_0, Y_0) \in \mathbb{R}^2$. Suppose that $\mathbb{F}$ satisfies the two conditions: $\mathbb{F}(X_0,Y_0) = Z_0$ and $\frac{d \mathbb{F}}{d Y}(X_0,Y_0) \neq 0$. Then there exist open intervals $U$ and $V$, with $X_0 \in U$ and $Y_0 \in V$, and a unique function $F : U \to V$ satisfying $\mathbb{F}[X, F(x)] = Z_0$ for all $X \in U$, and this function $F$ is continuously differentiable with: $\frac{dY}{dX}(Y_0) = F'(Y_0) = -\left [ \frac{\frac{d\mathbb{F}}{d X}(X_0, Y_0)}{\frac{d\mathbb{F}}{dY}(X_0, Y_0)} \right]$ ".

Is this the implicit function theorem? I'm having trouble understanding what this is. Does the unique function $F : U \to V$ mean the $g(x)$ that is usually referred to when talking about the IFT?

I don't understand the very last part of the theorem.

  • $\begingroup$ it is, and there are generalizations. In [ en.wikipedia.org/wiki/Implicit_function_theorem ] there are a good few examples $\endgroup$
    – janmarqz
    Jan 17, 2014 at 17:38
  • $\begingroup$ It may help to consider a linear $F$ to develop intuition. If $F(x,y) = ax+by$, and $F(x,y) = 0$, then we have $x = - { b \over a}y$, which corresponds to the derivative formula above (that is ${dx \over dy} = - {b \over a}$). $\endgroup$
    – copper.hat
    Jan 17, 2014 at 17:40

1 Answer 1


Here's the context. You have some equation $G(x,y)=C$, where $C$ is constant, that you'd like to solve for $y$ as a function of $x$. You require that $y$ vary in a smooth way as a function of $x$ (this is critical for uniquness because of things like +/- square root.) The assumption is that you've found one point $(x_{0},y_{0})$ where $G(x_{0},y_{0})=C$, and you want to solve for a smooth function $y=y(x)$ for $x$ near $x_{0}$, such that $y(x_{0})=y_{0}$.

This theorem gives you conditions under which you can solve for $y=y(x)$ as a function of $x$ such that $G(x,y(x))=C$. The only conditions are that $G$ must be continuously differentiable, and that $$ \left.\frac{\partial G}{\partial y}G(x,y)\right|_{x=x_{0},y=y_{0}} \ne 0. $$ You won't normally be able to solve the equation globally, but you can for at least $x$ close enough to $x_{0}$. The above condition is intuitively obvious because, if such $y$ exists, then $y'$ can be calculated from the chain rule: $$ 0=\frac{d}{dx}C=\frac{d}{dx}G(x,y(x))=\frac{\partial G}{\partial x}(x,y(x))+\frac{\partial G}{\partial y}(x,y(x))y'(x), $$ which leads to $$ y'(x) = -\left.\frac{\partial G}{\partial}(x,y(x))\right/\frac{\partial G}{\partial y}(x,y(x)). $$ So you need to at least know that $\frac{\partial G}{\partial y}(x_{0},y_{0})\ne 0$. And, that condition is enough in order to get a solution for all $x$ near $x_{0}$ with $y(x_{0})=y_{0}$ and $y$ smoothly varying with $x$. And, of course, the derivative $y'$ is correctly given by the above expression.


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