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Given a $ C^{\infty} $ function $ f:R^{2} \to R $ with $f(a,b)=0$. Suppose that $\left.\frac{df}{dy} \right| _{(a,b)}≠0 $ The implicit function theorem states that the level set $\{{(x,y):f(x,y)=0}\}$ is the graph of a smooth function $y=g(x)$ near $(x,y)=(a,b)$

I want to compute $ \left.\frac{dg}{dx} \right|_{a} $ and $ \left.\frac{d^2g}{dx^2}\right|_{a} $

I am studying for an exam. I have trouble understanding and solving the problems which are related to Implicit Function Theorem. How can we solve this problem? I need some more problems related to this theorem. Thanks in advance.

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Differentiate both sides of $f(x,y)=0$ and solve the resulting equation for $\left. \frac{dg}{dx}\right\vert _{a}$. –  Américo Tavares Mar 9 '12 at 21:17
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As Américo said, for $x$ near $a$ you have (note that $g(a)=b$) $$ f(x,g(x))=0. $$

Differentiating (using the chain rule) on the neighbourhood of $x$ given by the Implicit Function Theorem we get \begin{equation}\tag{1} f_x(x,g(x))+f_y(x,g(x))g'(x)=0. \end{equation} Evaluating at $x=a$ and solving, we get $$ g'(a)=-\frac{f_x(a,b)}{f_{y}(a,b)}. $$ To get the second derivative, we differentiate $(1)$ to get $$ f_{xx}(x,g(x))+f_{xy}(x,g(x))g'(x)+(f_{yx}(x,g(x))+f_{yy}(x,g(x)))g'(x)+f_y(x,g(x))g''(x)=0. $$ Evaluating at $x=a$ and solving, we get $$ g''(a)=-\frac{f_{xx}(a,b)+2f_{xy}(a,b)g'(a)+f_{yy}(a,b)g'(a)}{f_y(a,b)} $$

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