Derivative of implicite function.

Posted March 20th, 2008 by Isoscel

Let consider the equation $F(x,y)=x^2+y^2-1=0$
We know that the set of solution of this equation is the set of points in plane at distance 1 from the origin, or a circle of radius 1 with center in (0,0).
We want to give a description of this set depending on a single variable instead of two.
This is possible only local ,not for the whole set of solutions.
But the collection of local solutions can give us a complete information about the circle.
Let $(a,b)\in R^2$ with $a^2+b^2-1=0$ and $b\neq 0$
For $(x,y)$ in a small neighborhood of $(a,b)$ we have $y\neq0$
Implicit function theorem says that there is a local function f in a neighborhood of a with values in a neighborhood of b with f(a)=b and

$x^2+f^2(x)-1=0$

These function is derivable so

$2x +2f(x)f'(x)=0$

But as f has values in a neighborhood of $b\neq 0$ we can suppose $f(x)\neq 0$ so we can divide by f(x) and get

$f'(x)=-\frac{x}{f(x)}$

We can continue and have second derivative

$f"(x)=-\frac{f(x)-xf'(x)}{f^2(x)}=-\frac{f^2(x)+x^2}{f^3(x)}=-\frac{1}{f^3(x)}$

In this case there are two distinguished functions $f_{+}(x)=\sqrt{1-x^2}$ and $f_{-}(x)=-\sqrt{1-x^2}$ both verifying the above properties.
But what about equation

$F(x,y)=x^3+y^3-3xy=0$

It is possible to write again y=f(x) ?
Answer is given by the same Implicit function theorem but in this case is not so comfortable to use explicit solution.See Folium of Descartes

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Derivative of implicite function.

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