How Partial Derivative Changing Variable Formula works ?

[pyfgcr]

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

 

[tiny-tim]

welcome to pf!

hi pyfgcr! welcome to pf!

… 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 )

 

[pyfgcr]

But x is express in u and v, whereas in ∂u/∂x, y is kept constant.
What explain it ?

 

[HallsofIvy]

Again, that is the definition of the partial derivative.

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

When taking the derivative with respect to one variable, all other variables are held constant.

 

[pyfgcr]

So in (∂x/∂u)_y, by definition:
(∂x)/(∂u)_y=lim_{h\rightarrow0}\frac{x(u_{0},y_{0}+h)-x(u_{0},y_{0})}{h}
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.

 

[Muphrid]

On a realistic level, with the variables you've been given, you're going to need to invert something to get u = u(x,y) and the same for v.

 

[pyfgcr]

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

 

[tiny-tim]

hi pyfgcr!

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 ​

 

[pyfgcr]

Thanks a lot!