# Partial derivatives after a transformation

1. Nov 25, 2012

### MarkovMarakov

Suppose I have a transformation

$$(x'_1,x'_2)=(f(x_1,x_2), g(x_1,x_2))$$ and I wish to find $$\partial x'_1\over \partial x'_2$$ how do I do it?

If it is difficult to find a general expression for this, what if we suppose $f,g$ are simply linear combinations of $x_1,x_2$ so something like $ax_1+bx_2$ where $a,b$ are constants?

2. Nov 25, 2012

### arildno

Let the following transformation rules hold:
$$x'_1=f(x_1,x_2)$$
$$x'_2=g(x_1,x_2)$$
$$x_1=F(x'_1,x'_2)$$
$$x_2=G(x'_1,x'_2)$$

Then, we clearly have, for example the identity:
$$x'_1=f(F(x'_1,x'_2),G(x'_1,x'_2))$$
Since the pair of marked variables (and the unmarked) are independent from each other, we have the requirement on the functional relationship:
$$\frac{\partial{x}_{1}^{'}}{\partial{x}_{2}^{'}}=0=\frac{\partial{f}}{\partial{x}_{1}}\frac{\partial{F}}{\partial{x}_{2}^{'}}+\frac{\partial{f}}{\partial{x}_{2}}\frac{\partial{G}}{\partial{x}_{2}^{'}}$$

Last edited: Nov 25, 2012
3. Nov 25, 2012

### arildno

Suppose you have, say:
x=u^2-v^2
y=u+v
How can you utilize those implied requirements in the previous post to derive the the correct representations of u and v in terms of x and y?

Note that in this case, it is fairly trivial to find it by algrebrac means; dividing the first relationship by the second, we have x/y=u-v, and thus:
u=1/2(x/y+y), v=1/2(y-x/y)

However, using instead the four identities gained from the above requirements, we have
$$\frac{\partial{x}}{\partial{x}}=1=2u\frac{\partial{u}}{\partial{x}}-2v\frac{\partial{v}}{\partial{x}}$$
$$\frac{\partial{x}}{\partial{y}}=0=2u\frac{\partial{u}}{\partial{y}}-2v\frac{\partial{v}}{\partial{y}}$$
$$\frac{\partial{y}}{\partial{y}}=1=\frac{\partial{u}}{\partial{y}}+\frac{\partial{v}}{\partial{y}}$$
$$\frac{\partial{y}}{\partial{x}}=0=\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{x}}$$

from these, you should be able to derive the above relations.

Note, for example, that by combining the first and fourth and the known relation y=u+v, you get:
$$\frac{\partial{u}}{\partial{x}}=\frac{1}{2y}$$
meaning that we must have u(x,y)=x/2y+h(y), where h(y) is some function of y.
Applying the fourth, you get v(x,y)=b(y)-x/2y, for some b(y)

By using known and gained information, you will be able to determine h(y) and b(y) from the second and third relationships.

Last edited: Nov 25, 2012
4. Nov 25, 2012

### MarkovMarakov

Thank you, arildno.