# Partial derivative of function with respect to itself

I'm taking a physics course and I'm just getting familiar with their derivative notation, so I'm confused as to this notation:

Let $x_i = x_i(y_1,y_2) \quad i=1,2$ and I want to calculate

$$\frac{\partial(x_1,x_2) }{\partial(x_1,x_2) }=\det( \begin{matrix} \frac{\partial(x_1) }{\partial(x_1) } & \frac{\partial(x_1) }{\partial(x_2) } \\ \frac{\partial(x_2) }{\partial(x_1) } & \frac{\partial(x_2) }{\partial(x_2) } \\ \end{matrix})$$

So at first that's just the definition of the expression on the left, i.e. the determinant of the Jacobian, but what is for example ? $\frac{\partial(x_1) }{\partial(x_1) }$

Is $\frac{\partial(x_1) }{\partial(x_1) }= \frac{\partial(x_1) }{\partial(y_1) } \frac{\partial(y_1) }{\partial(x_1) } = \frac{\partial(x_1) }{\partial(y_1) } (\frac{\partial(x_1) }{\partial(y_1) })^{-1}= 1$ ? Via the chain rule

Or is it correct to just say $\frac{\partial(x_1,x_2) }{\partial(x_1,x_2) }= \frac{\partial(x_1,x_2) }{\partial(y_1,y_2) }\frac{\partial(y_1,y_2) }{\partial(x_1,x_2) }= \frac{\partial(x_1,x_2) }{\partial(y_1,y_2) }(\frac{\partial(x_1,x_2) }{\partial(y_1,y_2) })^{-1}=E_2$

How do I properly evaluate this with the chain rule?

• Your $x$ coordinates are given as functions of the $y$ coordinates. So aren't you after $\frac{\partial(x_1,x_2)}{\partial(y_1,y_2)}$? – Matthew Leingang Nov 11 '17 at 15:19

No need for the chain rule. In single-variable calculus, if $f(x)=x$ then $f'(x)=1$. Or, in other words, $dx/dx=1$. So $\partial x_1/\partial x_1=1$ in just the same way.