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Let $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. For each $x \in \mathbb{R}$, define a function $g_x: \mathbb{R} \rightarrow \mathbb{R}$ by $g_x(y)=f(x,y)$. Suppose that for each $x$, there is a unique y such that $g_x'(y)=0$; let $c(x)$ be this $y$.

Suppose the partial derivative $\frac{\partial^2 f}{\partial y^2} \neq 0$ for all $(x,y)$. Show that $c$ is a differentiable function and $$c'(x)=-\frac{\frac{\partial^2f}{\partial x \partial y}(x,c(x))}{\frac{\partial^2f}{\partial^2 y}(x,c(x))}$$

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Well, $ \frac{\partial g_x}{\partial y}(c(x)) = \frac{\partial f}{\partial y}(x,c(x)) = 0 $ for all $x$. Try to differentiate with respect to $x$ using the chain rule. After organising the terms you will see that it works out.

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Can you just start the differentiation. I am not sure how to proceed – Jean-Francois Rossignol Mar 31 '13 at 13:44
$0 = \frac{\partial }{\partial x} \frac{\partial f(x,c(x))}{\partial y} = \frac{\partial^2f }{\partial x\partial y}(x,c(x)) + \frac{\partial^2f }{\partial y^2} (x,c(x)) c'(x) $ – Henrik Finsberg Mar 31 '13 at 13:45
Just one more thing. How do I show that c is differentiable ? Can you add it in your solution perhaps – Jean-Francois Rossignol Mar 31 '13 at 13:49
@HenrikFinsberg You didn't show why c is differentiable – Carpediem Mar 31 '13 at 13:57
But this at least answers the OP's question partially. +1 – user1551 Mar 31 '13 at 14:01

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