Partial derivatives inverse question

Question

If $ \partial u/\partial v=a $, then $ \partial v/\partial u=1/a$?

Answer 1

Functions of a Single Variable

For functions of one variable, if $y=f(x)$ is strictly monotone and differentiable on an interval, and $f'(x)\ne 0$ in that interval, then the inverse function $x=f^{-1}(y)$ is also strictly monotone and differentiable in the corresponding interval and

$$\bbox[5px,border:2px solid #C0A000]{\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}}\tag 1$$

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EXAMPLE:

Suppose $y=\sin(x)$ for $x\in (-\pi/2,\pi,2)$. Note that the sine function is monotone and differentiable on $(-\pi/2,\pi/2)$ with $\frac{dy}{dx}=\cos(x)$ and $\cos(x)\ne 0$.

The inverse function, call it $x=\arcsin(y)$ for $y\in (-1,1)$, is therefore monotone and its derivative is

$$\frac{dx}{dy}=\frac{1}{\cos(x)}=\frac{1}{\sqrt{1-y^2}}$$

Therefore, we have $\frac{d\,\arcsin(y)}{dy}=\frac{1}{\sqrt{1-y^2}}$.

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Functions of a Two Variables

The relationship in $(1)$ does not apply, in general, to functions of more than one variable. As an example, examine the transformation of Cartesian coordinates $(x,y)$ to polar coordinates $(\rho,\phi)$ as given by

$$\begin{align} \rho &=\sqrt{x^2+y^2}\\\\ \phi &=\operatorname{arctan2}(y,x) \end{align}$$

and

$$\begin{align} x&=\rho \cos(\phi)\\\\ y&=\rho \sin(\phi) \end{align}$$

We examine the relationship between $\frac{\partial \rho }{\partial x}$ and $\frac{\partial x}{\partial \rho}$ to see if $(1)$ holds. Note that

$$\begin{align} \frac{\partial \rho }{\partial x}&=\frac{x}{\rho}\\\\ & =\cos(\phi)\\\\ &=\frac{\partial x}{\partial \rho} \end{align}$$

Therefore, $\frac{\partial \rho }{\partial x}\ne \frac{1}{\frac{\partial x}{\partial \rho}}$ and $(1)$ does not hold (unless $y=0$).

Instead of the relationship $(1)$ holding, we have instead

$$\begin{equation} \begin{pmatrix} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi} \end{pmatrix} \begin{pmatrix} \frac{\partial \rho}{\partial x} & \frac{\partial \rho}{\partial y} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation}$$

whereupon matrix inversion becomes

$$\begin{equation} \begin{pmatrix} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi} \end{pmatrix} =\begin{pmatrix} \frac{\partial \rho}{\partial x} & \frac{\partial \rho}{\partial y} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} \end{pmatrix}^{-1} \tag 2\end{equation}$$

Note that $(2)$ is the analog of $(1)$ and applies whenever a transformation and its inverse exists and is prescribed by differentiable functions. Moreover, it can be generalized to any number of variables.