Inverse function theorem for partial derivatives of a vector function

Question

I have a simple vector function:

$$\mathbf{y} = a\mathbf{x}\qquad a\in\mathbb{R},\;\mathbf{x},\mathbf{y}\in\mathbb{R}^n$$

The inverse is obviously:

$$\mathbf{y}^{-1} = {1\over a}\mathbf{x}$$

The inverse function theorem (at least for scalars) states that:

$$\left[f^{-1}(y)\right]'=\frac{1}{f'(f^{-1}(y))}$$

But the partial derivatives of $\mathbf{y}$ and $\mathbf{y}^{-1}$ with respect to $a$ are not related that way:

$$\begin{align}(\mathbf{y})'_a &= \mathbf{x} \\ (\mathbf{y}^{-1})'_a &= -{1\over a^2}\mathbf{x} \end{align}$$

What is the relation between the above derivatives, then?

Answer 1

The derivative of a function $F:\mathbb{R}^n\to \mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix

$$J_F(x) = \begin{pmatrix} \dfrac{\partial F_1}{\partial x_1}(x) & \cdots & \dfrac{\partial F_1}{\partial x_n}(x)\\ \vdots & \ddots & \vdots\\ \dfrac{\partial F_n}{\partial x_1}(x) & \cdots & \dfrac{\partial F_n}{\partial x_n}(x) \end{pmatrix}$$

And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :

$$J_{F^{-1}}(x) = \left(J_F(x)\right)^{-1}$$

You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem

Answer 2

The derivative w.r.t. $a$ is

$${\mathbb{d}\mathbf{y}\over \mathbb{d}a} = \mathbf{x}$$

and the derivative of inverse is the reciproval of above, i.e.

$${\mathbb{d}a\over \mathbb{d}\mathbf{y}} = {1\over \mathbf{x}}$$

The notation caused confusion about what is the inverse function. The inverse is function of $a$.