# I don't understand how partial derivatives work exactly

1. Nov 11, 2012

### platinumrice

what does d/ds (e^s cos(t)du/dx + e^s sin(t)du/dy) give, given that u = f(x,y)

i don't know how to manipulate d/ds and how to derive using d/ds. i am trying to simplify the expression, but i don't know, i just get stuck in the middle of can't get farther than here:

http://s12.postimage.org/s095nf6il/image.jpg [Broken]

i watched countless video tutorials, can do basic exercises, but i can't do this, please help me. there is a piece of the puzzle missing in my head.

Last edited by a moderator: May 6, 2017
2. Nov 11, 2012

### Vorde

Partial Derivatives are when you assume all but one of the variables are constants, and then derive with regards to whatever variable you chose to stay a variable. It can be confusing to get started with, but once you get used to it it'll get easy. Let's just look at a small part of that equation, $\frac{\partial}{\partial{s}}(e^s\cos{t})$

Now we have chosen s to be the variable we keep a variable, t becomes a constant. And if t is a constant, then so is $\cos{t}$. If you remember from single variable calculus, a constant multiplier can be brought outside of the derivative operator, so $$\frac{\partial}{\partial{s}}(e^s\cos{t})$$ becomes $$\cos{t}\frac{\partial}{\partial{s}}(e^s)$$
This is now easy to solve, the derivative of $e^s$ with respect to s is $e^s$, and so the whole thing becomes: $\cos{t}\cdot e^s$

Just look at it piece by piece, and remember all your other variables are going to become constants. If it helps try replacing things like t and $\cos{t}$ with constants like a and b just to remember you treat them like constants.

3. Nov 11, 2012

### platinumrice

i know that, but in this case, you have several derivatives.

4. Nov 11, 2012

### platinumrice

5. Nov 11, 2012

### chiro

Hey platinumrice and welcome to the forums.

In terms of why the others become constant, you need to consider that all variables are orthogonal to each other and if they aren't, then you need to resort to differential geometry and the calculus on manifolds.

But in an orthogonal system, the derivatives between two variables (call them x and y) have the property dx/dy = dy/dx = 0 for all values of x and y and thus no dependency exists between them.

This is the primary reason why we can treat things as a constant: It's because of the orthogonality criteria that makes the two variables independent.

Again if you are not in typical R^n space, you need to use calculus on manifolds and what this means is that the variables are inter-twined together.

The other to think about it is that for a certain fixed value of the other parameters, you fix those and instead of becoming variables (like x, y) they become constants (x0, y0) and then when you differentiate the function what you are doing is you treating this portion as a "slice" of the total function just like you cut a slice in a cake except this slice has zero width.

So the orthogonality condition and the slice idea can aid your understanding in why things are the way they are.

6. Nov 12, 2012

### HallsofIvy

You mean, of course, partial derivative, $\partial/\partial s$, $\partial u/\partial x$, and $\partial u/\partial y$.
Essentially, unless s and t depend upong x and y in some way you haven't told us, you have four independent variables, differentiating with respect to s, can treat t, x, and y as well as $\partial u/\partial x$ and $\partial u/\partial y$ as constants. That is, your function consists just of $e^s$ times a constant and, since the derivative of $e^x$ is just $e^x$ again, the derivative with respect to s of what you give is exacty itself again!

Last edited by a moderator: May 6, 2017