2.2 Partial derivative

Given $f:E\to {\mathbb{R}}^p$, ${\mathbf{x}}_0\in E\subseteq {\mathbb{R}}^m$ is the interior point, $\{{\mathbf{v}}_1,...,{\mathbf{v}}_m\}$ compose a set of bases in ${\mathbb{R}}^m$. Therefore

is called the coordinate based on ${\mathbf{v}}_1,...,{\mathbf{v}}_m$.

Let $\mathbf{v}=\sum _{i=1}^m{\xi }_i{\mathbf{v}}_i$, assuming $f$ is differentiable at ${\mathbf{x}}_0$, then

Mark

which is called the partial derivative of $f$ at ${\mathbf{x}}_0$ over the $i^{\text{th}}$ component under the coordinate system $(x_1,...,x_m)$. Define the coordinate-projection function $x_i:{\mathbb{R}}^m\to \mathbb{R}$,

It’s a linear function, therefore, $\text{d}{x}_i=x_i:{\mathbb{R}}^m\to \mathbb{R}$, $x_i:{\mathbb{R}}^m\to \mathbb{R}$,

is also linear. When $p=1$, we have

which is called the (total) differential of $f$ at ${\mathbf{x}}_0$. Here $\text{d}{x}_i$ are functions, not numbers! Take $z=f(x,y)$ as an example, we have

Consider $$\begin{array}{rl}\partial f({\mathbf{x}}_0)(\mathbf{v})& =\sum _{i=1}^m\frac{\partial f}{\partial x_i}({\mathbf{x}}_0){\xi }_i=\underset{\text{The representative matrix of }\partial f({\mathbf{x}}_0)}{\underbrace{\left(\begin{array}{ccc}\frac{\partial f}{\partial x_1}({\mathbf{x}}_0)& \cdots & \frac{\partial f}{\partial x_m}({\mathbf{x}}_0)\end{array}\right)}}\left(\begin{array}{c}{\xi }_1\\ ⋮\\ {\xi }_m\end{array}\right)\\ \text{d}f({\mathbf{x}}_0)& =\left(\begin{array}{ccc}\frac{\partial f}{\partial x_1}({\mathbf{x}}_0)& \cdots & \frac{\partial f}{\partial x_m}({\mathbf{x}}_0)\end{array}\right)\end{array}$$

When $p>1$, consider a mapping

Here

is called the Jacobi matrix of $f$ at ${\mathbf{x}}_0$.

Chain rule. $$\begin{array}{rl}\partial (G\circ F)({\mathbf{x}}_0)& =\partial G({\mathbf{y}}_0)\circ \partial F({\mathbf{x}}_0)\\ & \Updownarrow \\ \text{J}(G\circ F)({\mathbf{x}}_0)& =\text{J}G({\mathbf{y}}_0)\text{J}F({\mathbf{x}}_0)\end{array}$$

Assuming

then $$\begin{array}{rl}{\left(\frac{\partial z_i}{\partial x_j}\right)}_{l\times m}& ={\left(\frac{\partial z_i}{\partial y_k}\right)}_{l\times n}{\left(\frac{\partial y_k}{\partial x_j}\right)}_{n\times m}\\ \frac{\partial z_i}{\partial x_j}& =\sum _{k=1}^n\frac{\partial z_i}{\partial y_k}\cdot \frac{\partial y_k}{\partial x_j}\mathrm{\forall }i,j\end{array}$$

If $G=g$ is a function, i.e.

therefore $$\begin{array}{rl}\text{d}z& =\sum _{i=1}^m\frac{\partial (g\circ F)}{\partial x_i}\text{d}{x}_i=\sum _{i=1}^m\left(\sum _{k=1}^n\frac{\partial z}{\partial y_k}\frac{\partial y_k}{\partial x_i}\right)\text{d}{x}_i\\ & =\sum _{k=1}^n\left[\frac{\partial z}{\partial y_k}\sum _{i=1}^m\frac{\partial y_k}{\partial x_i}\text{d}{x}_i\right]=\sum _{k=1}^n\frac{\partial z}{\partial y_k}\text{d}{y}_k\\ \text{d}z& =\sum _{i=1}^m\frac{\partial z}{\partial x_i}\text{d}{x}_i=\sum _{k=1}^n\frac{\partial z}{\partial y_k}\text{d}{y}_k\end{array}$$

which is called the formal invariance of first-order derivative, meaning for any set of variables to express $z$, the form of the differential of $z$ remains invariant.

Example 2.2.1 Orthogonal coordinate and polar coordinate. Given $z=f(x,y)$,

Find the relation between $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}$ and $\frac{\partial g}{\partial r},\frac{\partial g}{\partial \theta }$. Notice that $$\begin{array}{rl}\text{d}z& =\frac{\partial z}{\partial x}\text{d}x+\frac{\partial z}{\partial y}\text{d}y=\frac{\partial f}{\partial x}\text{d}x+\frac{\partial f}{\partial y}\text{d}y\\ \text{d}z& =\frac{\partial z}{\partial r}\text{d}r+\frac{\partial z}{\partial \theta }\text{d}\theta =\frac{\partial g}{\partial r}\text{d}r+\frac{\partial g}{\partial \theta }\text{d}\theta \\ \text{d}x& =\frac{\partial x}{\partial r}\text{d}r+\frac{\partial x}{\partial \theta }\text{d}\theta =\cos \theta \text{d}r-r\sin \theta \text{d}\theta \\ \text{d}y& =\frac{\partial y}{\partial r}\text{d}r+\frac{\partial y}{\partial \theta }\text{d}\theta =\sin \theta \text{d}r+r\cos \theta \text{d}\theta \\ \frac{\partial g}{\partial r}& =\frac{\partial f}{\partial x}\cos \theta +\frac{\partial f}{\partial y}\sin \theta \\ \frac{1}{r}\frac{\partial g}{\partial \theta }& =\frac{\partial f}{\partial x}\sin \theta -\frac{\partial f}{\partial y}\cos \theta \\ \left(\begin{array}{cc}\frac{\partial z}{\partial r}& \frac{\partial z}{\partial \theta }\end{array}\right)& =\left(\begin{array}{cc}\frac{\partial z}{\partial x}& \frac{\partial z}{\partial y}\end{array}\right)\left(\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right)\end{array}$$

Proof Prove only $m=2$. $z=f(x,y)$, we need

• $\frac{\partial f}{\partial x}(a,b)$ exists.
• $\frac{\partial f}{\partial y}(x,y)$ exists on $U$ near $(a,b)$ and is continuous at $(a,b)$.

We use the 1-norm here, so it is needed to be proven that when $(x,y)\to (a,b)$,

$$\begin{array}{rlr}& f(x,y)-f(a,b)-\frac{\partial f}{\partial x}(a,b)(x-a)-\frac{\partial f}{\partial y}(a,b)(y-b)& \\ & =f(x,y)-f(x,b)-\frac{\partial f}{\partial y}(a,b)(y-b)& (1)\\ & +f(x,b)-f(a,b)-\frac{\partial f}{\partial x}(a,b)(x-a)& (2)\end{array}$$

Since $\frac{\partial f}{\partial x}(a,b)$ exists, for any $ϵ>0$, there exists ${\delta }_1(ϵ,a,b)>0$ such that

According to Lagrange’s intermediate theorem, $$\begin{array}{rlr}f(x,y)-f(x,b)& =\frac{\partial f}{\partial y}(x,\xi )(y-b)& \\ \xi & =(1-t(x,y))b+t(x,y)y& 0\le t(x,y)\le 1\\ (1)& =[\frac{\partial f}{\partial y}(x,\xi )-\frac{\partial f}{\partial y}(a,b)](y-b)& \end{array}$$

Since $\frac{\partial f}{\partial y}$ is continuous at $(a,b)$, for any $ϵ>0$, there exists ${\delta }_2(ϵ)>0$ such that

Here $|x-a|+|\xi -b|\le |x-a|+|y-b|<{\delta }_2$, so $|(1)|\le ϵ|y-b|$. Select $\delta =min\{{\delta }_1(ϵ),{\delta }_2(ϵ)\}$, then for any $|x-a|+|y-b|<\delta (ϵ)$,

Terminally $f$ is differentiable at $(x,y)$. $◻$