I don't understand how partial derivatives work exactly

In summary: 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
  • #1
platinumrice
5
0
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:
Physics news on Phys.org
  • #2
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
i know that, but in this case, you have several derivatives.
 
  • #5
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
platinumrice said:
what does d/ds (e^s cos(t)du/dx + e^s sin(t)du/dy) give, given that u = f(x,y)
You mean, of course, partial derivative, [itex]\partial/\partial s[/itex], [itex]\partial u/\partial x[/itex], and [itex]\partial u/\partial y[/itex].
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 [itex]\partial u/\partial x[/itex] and [itex]\partial u/\partial y[/itex] as constants. That is, your function consists just of [itex]e^s[/itex] times a constant and, since the derivative of [itex]e^x[/itex] is just [itex]e^x[/itex] again, the derivative with respect to s of what you give is exacty itself again!

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:

1. What is a partial derivative?

A partial derivative is a mathematical concept used to describe how a function changes in response to changes in one of its variables, while holding all other variables constant. It is denoted by ∂ (the partial derivative symbol) and is often used in multivariate calculus and physics.

2. How is a partial derivative calculated?

To calculate a partial derivative, you take the derivative of the function with respect to the specific variable you are interested in, treating all other variables as constants. This involves using the rules of differentiation, such as the power rule or product rule, and evaluating the resulting expression at a specific point.

3. What does a partial derivative represent?

A partial derivative represents the rate of change of a function with respect to a specific variable. It can also be thought of as the slope of the tangent line to a curve on a multi-dimensional graph at a specific point.

4. Why are partial derivatives important?

Partial derivatives are important in many fields, including mathematics, physics, engineering, and economics. They allow us to analyze how a system or process changes with respect to different variables, and are essential in optimizing functions and solving differential equations.

5. Can you give an example of a real-world application of partial derivatives?

One example of a real-world application of partial derivatives is in economics, where they are used to calculate marginal costs and marginal revenue in order to optimize production and pricing decisions. They are also used in physics to calculate rates of change in thermodynamic processes and in engineering to design efficient systems and structures.

Similar threads

  • Calculus
Replies
2
Views
2K
Replies
3
Views
1K
Replies
4
Views
1K
Replies
6
Views
1K
Replies
5
Views
974
Replies
2
Views
1K
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
902
Replies
25
Views
4K
Back
Top