graphical link between function and derivate

[mad]

Hello all,
first, excuse my english I dont speak it very well

I have a problem. We have two sheets. One are graphics of functions, and the other are graphics of the derivate of those function. Now my problem is I dont know how link a graphic of a function to the graphic of its derivate. I know that, for example, y = x^2, for ]-oo, 0[ , that the slope (sp?) will be negative. So why , on the graphic of the derivate which is y=2x, is the slope positive? How can I associate a function to its derivate?
thanks a lot

[mad]

If you dont understand what I'm asking, here is an exercice exactly like the one I'm talking about.
http://gmca.eis.uva.es/wims/wims.cgi?lang=es&+module=U1%2Fanalysis%2Fderdraw.en

Choose degree 3 or more and it asks to draw its derivate graphic. But I dont know how

[gazzo]

The gradient function of y = x^2, that is: y = 2x, is negative for values of x less than zero. So, although the slope of the gradient function is positive, you can see that value of the gradient function at say, x = -2 still gives the slope of y = x^2 at x = -2.

The slope of the gradient function would only be negative if the original function was y = -x^2.

It helps to plot the two graphs y = x^2 and y = 2x above and below each other, and matching respective x values on both, to get a feel for what's happening in the gradient function.

[kreil]

The relationship is:

When the function is increasing, the derivative is positive

When the function is decreasing, the derivative is negative

When the function is changing direction, the derivative is zero

So, in the case of f(x)=x^2 from -\infty\rightarrow 0 the function is decreasing and the derivative is negative. At the point (0,0) the function changes direction, so the derivative is zero, and from 0\rightarrow\infty the function is increasing so the derivative is positive.

[DeadWolfe]

Although the slope of 2x is postive, the value of the function is negative.

It is the value of the function which you must be concerned with.