# What is the funciton of $dx'^2$?

Let's say same point in two co-ordinate system has the following relation from partial derivatives,

$$dx'=\frac{\partial x'}{\partial x} dx + \frac{\partial x'}{\partial y} dy$$ and $$dy'=\frac{\partial y'}{\partial x} dx + \frac{\partial y'}{\partial y} dy.$$

I have to derive the conditions for $$dx'^2 + dy'^2 \propto dx^2 + dy^2.$$

Now my question is how do define $$dx'^2$$? My memory and hunch says, it should be $$dx'^2 = \left( \frac{\partial x'}{\partial x} \right)^2 dx^2 + 2 \frac{\partial x'^2}{\partial x \partial y} dxdy + \left( \frac{\partial x'}{\partial y} \right)^2 dy^2$$

However, I'm not sure. Is it correct?

Also, could anyone tell me a good text to revise these stuffs from. I did them a long time ago and forgot.

• The expression you wrote is not $dx'^2$ but $d^2x'$, the second differential of $x'$ as a function of $x$ and $y$. – GReyes Feb 24 at 7:26
• What does $\alpha$ mean? – William Elliot Feb 24 at 9:44
• It is not $\alpha$, it is $\propto$ = proportional to. – ponir Feb 24 at 18:18
• @GReyes, In that case what is the expression for $dx'^2$? – ponir Feb 24 at 18:19
• Actually, what you have is neither of them (just did not notice that you have the squares, not the pure second derivatives in your expression). $dx'^2$ would simply be what you have, replacing the mixed second derivative by $(dx'/dx)(dx'/dy)$ (you just square your expression for $dx'$). – GReyes Feb 24 at 23:31