Help me understand sum involving partial derivative

Question

We have a bijective function $f$: $ \mathbb{R}^n \rightarrow \mathbb{R}^n$ and its inverse $f^{-1}$. My textbook says:

$$ \sum_{k = 1}^n \frac{\partial f_i^{-1}}{\partial x_k}\cdot\frac{\partial f_k}{x_j} = \delta_{i,j} $$

How is this derived?

Answer 1

Well, you have that $$(f^{-1}\circ f)(x) = x.$$

So compute $$\frac{d}{dx_j} (f^{-1}\circ f)_i$$ in two ways: first, clearly $$\frac{d}{dx_j} x_i = \delta_{ij}.$$ Second, compute the derivative $$\frac{d}{dx_j}(f^{-1}\circ f)_i = \frac{d}{dx_j} f^{-1}_i[f(x)]$$ using the chain rule. What do you get?