Skip navigation.
Home

Derivative of implicite function.

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

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

This 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

Average: 1.5 (2 votes)