How Partial Derivative Changing Variable Formula works ?

  • Thread starter pyfgcr
  • Start date
  • #1
22
0

Homework Statement


The changing variable formula in partial derivative
f(u,v)
x=x(u,v)
y=y(u,v)
(∂f/∂x)y = (∂f/∂u)v(∂u/∂x)y + (∂f/∂v)u(∂v/∂x)y
I khow the how chain rule works, but I don't know why in the (∂f/∂u) v is constant and in the (∂u/∂x) y is constant

Homework Equations


The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,836
251
welcome to pf!

hi pyfgcr! welcome to pf! :smile:
… I don't know why in the (∂f/∂u) v is constant and in the (∂u/∂x) y is constant

that's what a partial derivative is

it's defined as being calculated with all the other variables kept constant

so it actually depends on what the other variables are: for example, if a function f is expressed both as f(x,y,z) and f(r,z,θ) (ie cartesian coordinates and cylindrical coordinates),

then, even though the z is the same, ∂f/∂z is different in each case

(in practice you'll probably avoid confusion by using different letters for the function, f(x,y,z) and g(r,z,θ) … but if you don't, you will need to write either (∂f/∂z)x,y or (∂f/∂z)r,θ :wink:)
 
  • #3
22
0
But x is express in u and v, whereas in ∂u/∂x, y is kept constant.
What explain it ?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
965
Again, that is the definition of the partial derivative.

The partial derivative, with respect to x, of f(x,y) at [itex](x_0, y_0)[/itex] is defined as
[tex]\lim_{h\to 0}\frac{f(x_0+h, y_0)- f(x_0,y_0)}{h}[/tex]
and the partial derivative with respect to y is
[tex]\lim_{y\to 0}\frac{f(x_0, y_0+h)- f(x_0,y_0)}{h}[/tex]

When taking the derivative with respect to one variable, all other variables are held constant.
 
  • #5
22
0
So in (∂x/∂u)y, by definition:
(∂x)/(∂u)y=[itex]lim_{h\rightarrow0}[/itex][itex]\frac{x(u_{0},y_{0}+h)-x(u_{0},y_{0})}{h}[/itex]
A little awkward, since x=x(u,v), I think

In equation: (∂f/∂x)y = (∂f/∂u)v(∂u/∂x)y + (∂f/∂v)u(∂v/∂x)y
Why in the term (∂f/∂u)v, v is kept constant, not y.
 
  • #6
834
2
On a realistic level, with the variables you've been given, you're going to need to invert something to get [itex]u = u(x,y)[/itex] and the same for [itex]v[/itex].
 
  • #7
22
0


that's what a partial derivative is

it's defined as being calculated with all the other variables kept constant

And in (∂u/∂x)y, with x, "the other variable" than u is v, but y is kept constant, not v
 
  • #8
tiny-tim
Science Advisor
Homework Helper
25,836
251
hi pyfgcr! :smile:
f(u,v)
x=x(u,v)
y=y(u,v)
And in (∂u/∂x)y, with x, "the other variable" than u is v …

no

x and y are functions of u and v

u and v are functions of x and y

in particular, u is a function of x and y

so, for u, the "other variable" than x is y

so ∂u/∂x keeps y constant :wink:
 
  • #9
22
0
Thanks a lot!
 

Related Threads on How Partial Derivative Changing Variable Formula works ?

Replies
5
Views
784
Replies
8
Views
2K
Replies
3
Views
6K
Replies
14
Views
10K
  • Last Post
Replies
1
Views
2K
Replies
14
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
781
Top