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?
Best Answer
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.