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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:

  1. Apply the implicit function theorem to obtain a differentiable local solution to $D_2f(x,y) = 0$ at each $(x, c(x))$.
  2. 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 $c$.
  3. 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?

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The notation $D_2f$ means the partial derivative of $f$ w.r.t. its second argument. $D_{2,1}f$ means $D_1(D_2f)$, and likewise with $D_{2,2}f$. –  Brian Bi May 11 '12 at 18:08
What convention does Spivak use? Some authors write differentiable and mean smooth i.e. $C^{\infty}$. –  Olivier Bégassat May 11 '12 at 18:55
@OlivierBégassat : I'm fairly sure that "differentiable" has its usual meaning in this chapter (it begins to mean "smooth" in chapter 5). –  Brian Bi May 11 '12 at 19:01
There could be a mistake in the problem as it is formulated. You should assume the function to be smooth enough. The solution obviously does. –  Olivier Bégassat May 11 '12 at 19:19

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