# Partial derivative of coordinates with respect to function

Let $$f : \mathbb{R}^n \rightarrow \mathbb{R}^n$$. Then

$$\frac{\partial f^i}{\partial x^j} = (\nabla f)^i_j$$

where $$\nabla f$$ is the Jacobian matrix of $$f$$. When reading this paper I came across the expression

$$\frac{\partial x^i}{\partial f^j}$$

Should I interpret this as

$$\frac{\partial x^i}{\partial f^j} = ((\nabla f)^{-1})^i_j$$

Imagine you can invert the problem $$x^i = x^i(f)$$. Clearly

$$x^i = x^i(f^1(x),\cdots,f^n(x))$$

Now apply the chain rule

$$\frac{\partial x^i}{\partial x^j} = \frac{\partial x^i}{\partial f^k} \frac{\partial f^k}{\partial x^j} = (\nabla_f x)^{i}_{\;k}(\nabla_x f)^{k}_{\;j} = \delta^i_j$$

That means that

$$\mathbb{1} = (\nabla_f x) (\nabla_x f)$$

Or in other words

$$\nabla_f x = (\nabla_x f)^{-1}$$

• Should be $\delta^i_j$ in your second equation, right? And dot product in the third? – user76284 Jan 12 at 0:20
• @user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product – caverac Jan 12 at 0:24
• Makes sense. I just always write the matrix product with a dot to make it clear I'm performing tensor contraction, and to distinguish it from the tensor product. – user76284 Jan 12 at 0:25