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

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

  • $\begingroup$ Should be $\delta^i_j$ in your second equation, right? And dot product in the third? $\endgroup$ – user76284 Jan 12 at 0:20
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    $\begingroup$ @user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product $\endgroup$ – caverac Jan 12 at 0:24
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    $\begingroup$ 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. $\endgroup$ – user76284 Jan 12 at 0:25

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