Let $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ be differentiable. For each $x \in \mathbb{R}$ define $g_x:\mathbb{R}\to\mathbb{R}$ by $g_x(y) = f(x,y)$. Suppose that for each $x$ there is a unique $y$ with $g_x'(y) = 0$; let $c(x)$ be this $y$.
(a) If $D_{2,2}(x,y) \neq 0$ for all $(x,y)$, show that $c$ is differentiable and $c'(x) = -\dfrac{D_{2,1}f(x,y)}{D_{2,2}f(x,y)}$.
Hint: $g_x'(y) = 0$ can be written $D_2f(x,y) = 0$.
I looked at the solution for this problem and it basically seems to go like this:
- Apply the implicit function theorem to obtain a differentiable local solution to $D_2f(x,y) = 0$ at each $(x, c(x))$.
- Since there is only one $c(x)$ for which $D_2f(x,c(x)) = 0$ at each $x$, each of these local solutions must agree with $c$. Therefore, by gluing together all these local solutions, we get an everywhere differentiable function identical to $c$.
- Since we know $c$ is a differentiable function, we can differentiate the relation $D_2f(x,c(x)) = 0$ using the chain rule (Theorem 2-9), which immediately gives the desired result.
I understand everything in this solution except for the following: before we can apply the implicit function theorem to $D_2f(x,y) = 0$, we need $D_2f$ to be continuously differentiable, and the problem statement doesn't seem to permit that assumption. How is this resolved?
Best Answer
This is a known oversight: http://www.jirka.org/spivak-errata.html
Assume that $f$ is twice continuously differentiable.