Computing derivatives of the implicit function

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

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 $$\tag{1} f_x(x,g(x))+f_y(x,g(x))g'(x)=0.$$ 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 $$g''(a)=-\frac{f_{xx}(a,b)+2f_{xy}(a,b)g'(a)+f_{yy}(a,b)g'(a)}{f_y(a,b)}$$