# Partial Derivatives?

1. Mar 3, 2008

1. The problem statement, all variables and given/known data

Okay, so at my school, you can get into Diff EQs without taking Calculus 3. So, I get most of the basis of it, but some things I am missing the boat on.

How does one go from

$d(\frac{1}{3}x^3y^3)=x^2y^3dx+x^3y^2dy$ ?

How do you carry out that operator? It says to take the derivative of that 2 variable function, but does not say with respect to what?

I am assuming it is some sort of partial derivative...but I am not sure?

Thanks

2. Mar 3, 2008

### quasar987

By definition,

$$df(x,y)=f_x(x,y)dx+f_y(x,y)dy$$

and df(x,y) is calles the total differential of f.

3. Mar 3, 2008

What does $f_x(x,y)dx$ mean? What is f_x?

4. Mar 3, 2008

### Pythagorean

I've just begun to accept this formula myself.

The d(stuff) on the left side of the equation contains two variables, x and y. so you take the partial derivative with respect to each of them and add them together.

f_x simply means the partial function's partial with respect to x. Test it out on the function you've given us.

5. Mar 3, 2008

Word.

So I take the derivative of $\frac{1}{3}x^3y^3$ first wrt x and then wrt y holding y then x constant, respectively, and then sum the results?

6. Mar 3, 2008

So I get $$y^3x^2\frac{df}{dy}+x^3y^2\frac{df}{dx}$$ then how do I cope with

the notations? How do I make it look like $x^2y^3dx+x^3y^2dy$?

7. Mar 3, 2008

### quasar987

You seem to have misunderstood slightly. I re-write what I wrote in post #2:

By definition,

$$df(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$

Here,

$$f(x,y)=\frac{1}{3}x^3y^3$$

8. Mar 3, 2008

I understand how to take the partial of $\frac{1}{3}x^3y^3$ wrt y and x, but I do not understand the NOTATION at all.

df(x,y) means to take the derivative of f wrt x (hold y as constant) and take the derivative of f wrt y (hold x constant), but what are the resulting notations?

9. Mar 3, 2008

### Pythagorean

it's like when you do a substitution in Calculus II.

you have some integral, like

$$\int{\frac{x }{1+x^2}dx}$$

so you use the substitution $$u = x^2$$

but you have to get the differential for u too, so you take the derivative of both sides of the above equation:

$$du = 2x dx$$ (here we have a function u = u(x), where our function above has two variables, so we require the partial derivative formula that quasar provided)

you could also see it as

$$\frac{du}{dx} = 2x$$

where you have multiplied by the differential dx, afterwards, treating it like an algebraic quantity (which we can do with practically all differentials in physics... might want to be more careful for a math teacher... maybe justify it)

Last edited: Mar 3, 2008
10. Mar 3, 2008

### Marco_84

think about it like in one variable... it is the increment of the function f due to the little dx and the "slope" is f'....right... now u have to coordinate axis x and y...

more formally the df is a matrix when the function is from R^n----->R^m

ciao

11. Mar 3, 2008

Okay. Let's keep it simple here guys.

12. Mar 3, 2008

### quasar987

You say you're confused with the notation.

But you are probably already familiar with the notion of a differential for a function of one variable? Given f differentiable, we define its differential as the function of two variables

df(x,dx) = f '(x)dx.

It says "given a value of x and an increment dx, the approximate resulting increment of f is df(x,dx)."

This is the same thing for the function of two variables. Given a point (x,y) and increments dx and dy in the x and y directions respectively, df(x,y,dx,dy) keeps tract of the approximate resulting increment of f based on the rates of change of f at (x,y) (the partial derivatives)

13. Mar 3, 2008

### Dick

Try it this way. Suppose x and y were both functions of t. x(t), y(t). Then d/dt(x^3*y^3/3)=x^2*y^3*dx/dt+x^3*y^2*dy/dt. Now suppose they are functions of q, x(q) and y(q). Same thing but with q substituted for t. Big deal. So let's just get it over with and say d(x^3*y^3/3)=x^2*y^3*dx+x^3*y^2*dy. We can divide by d(whatever) when we figure out what the 'whatever' is. It's just a notational thing.

14. Mar 4, 2008

:rofl: let's just get it over with... Wonderful. This makes sense. Thanks Dick.

15. Mar 4, 2008

### HallsofIvy

Staff Emeritus
I think you should go to the department chair and tell him/her how foolish it is to let people take partial differential equations without having taken multi-variable calculus first!

16. Mar 4, 2008

I never said that I was taking Partial Differential Equations. I am taking Differential Equations. And while defining exactness, some partial derivatives come up.

I just want to understand their proof fully.

17. Mar 4, 2008

### HallsofIvy

Staff Emeritus
I would consider both Calculus III (Calculus of several varibles) and Linear Algebra necessary prerequisites for "Ordinary Differential Equations".

18. Mar 4, 2008