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I have troble to understand derivative of vector.

In scalar case $y=f(x)$, the follow is truth

$$\frac{dy}{dx}=\left(\frac{dx}{dy}\right)^{-1}$$

In vector case, $\mathbf{y}=(y_1,y_2)$, $\mathbf{x}=(x_1,x_2)$

$\mathbf{y}=f(\mathbf{x})$

$$\frac{d\mathbf{y}}{d\mathbf{x}}=\left(\frac{d\mathbf{x}}{d\mathbf{y}}\right)^{-1}$$

My problem is why the follow does not hold? $$\frac{dy_1}{dx_1} = \left(\frac{dx_1}{dy_1}\right)^{-1}$$ Is there any demonstration can help me understand that? Thanks for your time.

Xu Hui
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2 Answers2

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You're inverting a matrix. In general, $(M^{-1})_{11}\ne(M_{11})^{-1}$. In terms of partial derivatives, the issue is they're defined with different variables held constant. While $\partial y_1/\partial x_1$ holds $x_i$ constant for $i\ne1$, $\partial x_1/\partial y_1$ holds $y_i$ constant for $i\ne1$.

J.G.
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  • Thanks sir, I accept that in general $(M^{-1}){11} \neq (M{11})^{-1}$. Is there any geometry demonstration why it is not equal? – Xu Hui Jul 15 '20 at 07:30
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    @XuHui Since the linear transformations matrices characterize are a special case of coordinate transformations, a geometric explanation would reduce to my point about partial derivatives. – J.G. Jul 15 '20 at 08:43
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Aspired by sir J.G., I now have a rough idea how to understand the problem.

$d\mathbf{x}$ is a vector, it has a direction and magnitude.

$d\mathbf{y}$ is another vector, is has another direction and mangnitude.

First we what calulate $d\mathbf{y}/d\mathbf{x}$, suppose $$\frac{d\mathbf{y}}{d\mathbf{x}}=\mathbf{M}$$ thus $$d\mathbf{y}=\mathbf{M}d\mathbf{x}$$

we can see that $\mathbf{M}$ is a mapping from $d\mathbf{x}$ to $d\mathbf{y}$. It leads to the rotation and deformation on the $d\mathbf{x}$.

Then we want calculate the $d\mathbf{x}/d\mathbf{y}$. Suppose $$\frac{d\mathbf{x}}{d\mathbf{y}}=\mathbf{Q}$$

we can see that $\mathbf{Q}$ is a mapping from $d\mathbf{y}$ to $d\mathbf{x}$. It leads to the rotation and deformation on the $d\mathbf{y}$.

thus $$d\mathbf{x}=\mathbf{Q}d\mathbf{y} = \mathbf{QM}d\mathbf{x}$$

$$ \mathbf{QM} = \mathbf{I}$$ $$ \mathbf{Q} = \mathbf{M}^{-1}$$

thus $\mathbf{Q}$ is a mapping can undo the effect of $\mathbf{M}$, that is the invert matrix of $\mathbf{M}$.

So the "inverse derivate of vector" should be a invert matrix $\mathbf{M}^{-1}$ but not the matrix organized by invert component $(M_{ij}^{-1})$.

Xu Hui
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