# How Partial Derivative Changing Variable Formula works ?

1. Jul 26, 2012

### pyfgcr

1. The problem statement, all variables and given/known data
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
2. Relevant equations
3. The attempt at a solution

2. Jul 26, 2012

### tiny-tim

welcome to pf!

hi pyfgcr! welcome to pf!
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 )

3. Jul 26, 2012

### pyfgcr

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

4. Jul 26, 2012

### 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.

5. Jul 26, 2012

### 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.

6. Jul 26, 2012

### 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 $v$.

7. Jul 27, 2012

### pyfgcr

Re: welcome to pf!

And in (∂u/∂x)y, with x, "the other variable" than u is v, but y is kept constant, not v

8. Jul 27, 2012

### tiny-tim

hi pyfgcr!
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

9. Jul 27, 2012

### pyfgcr

Thanks a lot!