# Partial differentiation

1. Feb 25, 2006

### gulsen

Say, E is dependent to x,y,z. I'm expecting it's derivative at $$x_0,y_0,z_0$$ to be

$$dE = \lim_{\substack{\Delta x\rightarrow 0\\\Delta y\rightarrow 0\\\Delta z\rightarrow 0}} E(x_0+\Delta x, y_0+\Delta y,z_0+\Delta z) - E(x_0,y_0,z_0)$$

But with following definition, it's not the thing above:

$$dE = \lim_{\substack{\Delta x\rightarrow 0\\\Delta y\rightarrow 0\\\Delta z\rightarrow 0}} \frac{E(x_0+\Delta x, y_0,z_0) - E(x_0,y_0,z_0)}{\Delta x} \Delta x + \frac{E(x_0, y_0+\Delta y,z_0) - E(x_0,y_0,z_0)}{\Delta y} \Delta y + \frac{E(x_0, y_0,z_0+\Delta z) - E(x_0,y_0,z_0)}{\Delta z} \Delta z = \frac{\partial E}{\partial x}dx + \frac{\partial E}{\partial y}dy + \frac{\partial E}{\partial z}dz$$.

Now, which is correct? (and why?!?)

Last edited: Feb 25, 2006
2. Feb 25, 2006

### HallsofIvy

Neither one. The differential (not "derivative") is
$$dE= \left(\lim_{\Delta x\rightarrow0}\frac{E(x_0+\Delta x,y_0,z_0)-E(x_0,y_0,z_0)}{\Delta x}\right)dx+ \left(\lim_{\Delta y\rightarrow 0}\frac{E(x_0,y_0+\Delta y,z_0)-E(x_0,y_0,z_0)}{\Delta y}\right)dy+ \left(\lim_{\Delta z\rightarrow 0}\frac{E(x_0,y_0,z_0+\Delta z)-E(x_0,y_0,z_0)}{\Delta z}\right)dz$$
That's the same as your second formula except for the "d" rather than "$\Delta$" in the numerator.

Why would you "expect" the first? In one variable, would you "expect" the differential of f(x) to be
$$df= lim_{\substack{\Delta x\rightarrow 0}}f(x_0+\Delta x)- f(x_0)$$
If f is any continuous function, that's 0!

Last edited by a moderator: Feb 25, 2006
3. Feb 25, 2006

### gulsen

Whoops, just a typo anyway.

That was just a guesswork based on intuition and my ignorance, I simply don't know. So tell me, why would you expect the latter?

4. Feb 26, 2006

### gulsen

Well, at least you could point some links.

5. Feb 26, 2006

### HallsofIvy

Or you could look at the definition of the differential in any calculus book:

$$df(x,y,z)= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz$$
which, if you go back to the definition of the partial derivatives, is the formula I gave.

6. Feb 26, 2006

### gulsen

Whoops, we've got a communication problem here, it seems.
I know that this is the definition, and I've written it before you as well. My question is: why? What geometry/mathematical quantity does this correntponds to?

7. Mar 1, 2006

### gulsen

From what I've found, it should be like this:
$$df(x,y,z)= \frac{\partial f}{\partial x}dx \vec{i} + \frac{\partial f}{\partial y}dy \vec{j} + \frac{\partial f}{\partial z}dz \vec{k}$$
But this way, for the magnitude of this df, we gotta multiply (dot product) df with itself and square root...

8. Mar 1, 2006

### jackiefrost

Just picture f as a function of two variables instead of three; z = f(x,y). Thus f can be represented by a surface S in 3 space. The total differential df in the equation
$$df(x,y)= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy$$
would represent the change in the z component of the tangent plane T of f at a point P at coordinates (x,y,z) as we go to some other point P' at (x+delta x,y+delta y,f(x+delta x,y+delta y)). In this case, since delta x and delta y are independent variables, we can say dx=delta x and dy=delta y; i.e dx and dy can be actual distance values. Notice that df is related to the z component of points in T but delta f is related to the z component of points in the surface that f generates.

It's even easier to picture if you just consider f as function of one variable: y=f(x). Though you lose the "partial" nature of the derivitive, the geometric picture you were looking for is still there. For example, you can picture some curved line that f generates and some tangent line T at a point P at (x, f(x)). If you take another point on the curve P' at (x+delta x, f(x+delta x)), dy would represent the verticle distance between a horizontal line through P and the tangent line T. In other words, dy would represent the distance that T changes in the y direction for some change dx=delta x in the x direction. We could imagine an animal called delta y (as distinct from dy) that represents the distance from the horizontal line through point P to the CURVE (rather than the tangent line T).

Why would I even begin to set here and type all that??? :surprised