by flyingpig
Tags: derivatives, partial
P: 2,568
1. The problem statement, all variables and given/known data

Let's say I have a function for a circle

$$x^2 + y^2 = C$$ where C is a constant.

Then this is a cylinder with the z-axis.

Now in my ODE book, we would normally define it as

$$F(x,y) = C = x^2 + y^2$$ as a level surface.

Now my question is about what the partial derivative with respect to x mean as opposed to (single-variable calculus) derivative with respect to x mean. Am I losing anything if I take one derivative over the other?

I should mention that many of these problems assume that F(x,y(x)).

$$\frac{\partial F}{\partial x} = 2x$$

$$\frac{\partial F}{\partial y} = 2y$$

$$\frac{\mathrm{d} F}{\mathrm{d} x} = 2x + 2y\frac{\mathrm{d} y}{\mathrm{d} x} = 0$$

So now my question is, what exactly is this
 $$2x + 2y\frac{\mathrm{d} y}{\mathrm{d} x} = 0$$
as opposed to
 2x
 PF Patron Sci Advisor Thanks Emeritus P: 38,424 $$2x+ 2y\frac{dy}{dx}$$ is the rate of change of the function $f(x, y(x))= x^2+ y(x)^2$ with respect to x- it measures how fast f(x,y(x)) changes as x changes. Of course that will depend upon exactly how y(x) changes as x changes- and that is what dy/dx tells you. Suppose y(x)= 3x. Then $d(x^2+ y^2)/dx= 2x+ 2y dy/dx= 2x+ 2y(3)= 2x+ 2(3x)(3)= 20x$.That is exactly the same as if you had replaced y with 3x from the start: $x^2+ (3x)^2= x^2+ 9x^2= 10x^2$ so $df/dx= 20x[/tex] P: 2,568  Quote by HallsofIvy $$2x+ 2y\frac{dy}{dx}$$ is the rate of change of the function [itex]f(x, y(x))= x^2+ y(x)^2$ with respect to x- it measures how fast f(x,y(x)) changes as x changes.
I thought that was what the partial derivative with respect to x is

PF Patron
Thanks
Emeritus
P: 38,424

No, the partial derivative of f with respect to x is the rate of change as x change assuming y does not change.
P: 1,668
 Quote by flyingpig I should mention that many of these problems assume that F(x,y(x)).
Right.

given a small change in x, dx, the change is x^2 is close to 2xdx and the change in y^2 is close to 2ydy. But dy = (dy/dx)dx.

If y were independent of x then dy/dx would be zero. This would just mean that x can change without a change in y. But the constraint C = x^2 + y^2 tells you that y is a function of x, at least locally.
P: 2,568
 Quote by lavinia Right. given a small change in x, dx, the change is x^2 is close to 2xdx and the change in y^2 is close to 2ydy. But dy = (dy/dx)dx. If y were independent of x then dy/dx would be zero. This would just mean that x can change without a change in y. But the constraint C = x^2 + y^2 tells you that y is a function of x, at least locally.
If I were to graph all three of those "derivatives" what would they look like? How do yuo even graph Fx alone?
 P: 2,568 Actualyl could I get a unit representation? Say F(x,y) was a position function of time (perhaps x(t) = x and y = y(t)) what would the different derivatives tell me?

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