Really need help with implicit functions

[newatthis]

hello

FIRST:
I thought I knew what implicit function was untill I saw this definition in my textbook:

Explicit function is y = f(x) ... ok
Implicit function is f(x, y) = 0 ... what?!!!!

I know that explicit function has one variable isolated( y = f(x) ), while implicit doesn't. But what does f(x, y) = 0 represent or mean?




SECOND:

Can you explain to me the formula for derivative of implicit function ( circle, ellipse, hyperbola )? The main point of this question is that I want to know why it works( not just how it works )!

EDIT: For example, if we have an implicit function with one term being y^2, then we must find the derivative of y^2 with respect to x, which is 2yy'.

But why would the correct way to be to find the derivative of y^2 with respect to x ?
Afteral implicit function is "different" from explicit function, so why would same rules ( like finding derivatives with respect to x ) apply here?

thanx in advance

 

[skeeter]

here are two very simple functions that I selected in order to show you the format f(x,y) = 0 versus y = f(x)

implicit function ...
xy - 1 = 0

same function expressed explicitly ...
y = 1/x

implicit function ...
x + y - 1 = 0

same function expressed explicitly ...
y = -x + 1

not all implicit relationships yield functions ...

the relation y^2 + y^2 = 1 (you should know this as the unit circle)
can actually be manipulated into two explicit functions ...
y = sqrt(1 - x^2) and y = -sqrt(1 - x^2), the upper and lower semicircles.

some implicit relations are rather difficult to express explicitly ...
x^2*y + y^2*x - 1 = 0

y = [-x^2 + sqrt(x^4 + 4x)]/(2x) and y = [-x^2 - sqrt(x^4 + 4x)]/(2x)

and some are almost impossible ...
x^5*sqrt(y) + y^2 - x^2*y + cos(x+y) - 5 = 0

when taking the derivative (w/r to x) of an implicit relation (in terms of x and y), you treat y as a function of x ...

d/dx[x^2 + y^2 - 1 = 0]
2x + 2y(dy/dx) = 0
dy/dx = -x/y

look in your text and on the web for other examples.