derivative

Prerequisites

• 向量空间
• funct-limit

Definitions

• real vector space: $<\mathbb{R},{\mathbb{R}}^n>$

real vector space can define inner product, which can induce norm, which can induce metric, which can induce topology.

• Multivariate real-valued function $f:{\mathbb{R}}^n\to \mathbb{R}$

• Multivariate real vector-valued function $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$

• Differentiability at a point Let $X$ be a subset of ${\mathbb{R}}^n$, and let $x_0\in X$ be an element of $X$ which is also a limit point of $X$. Let $f:X\to {\mathbb{R}}^m$ be a function. Let $L$ be a linear transformation from ${\mathbb{R}}^n$ to ${\mathbb{R}}^m$. We say that $f$ is differentiable at $x_0$ on $X$ with derivative $L$ and write $f^{\prime }(x_0):=L$ if

  $$\underset{x\to x_0,x\in X\setminus \{x_0\}}{\lim }\frac{d(f(x),f(x_0)+L(x-x_0))}{d(x,x_0)}=0$$

Uniqueness of derivatives Let $E$ be a subset of ${\mathbb{R}}^n$, $f:E\to {\mathbb{R}}^m$ be a function, $x_0\in E$ be an interior point of $E$, and let $L_1:{\mathbb{R}}^n\to {\mathbb{R}}^m$ and $L_2:{\mathbb{R}}^n\to {\mathbb{R}}^m$ be linear transformations. Suppose that $f$ is differentiable at $x_0$ with derivative $L_1$, and also differentiable at $x_0$ with derivative $L_2$. Then $L_1=L_2$.

proof:
because $L_1\ne L_2$, so there exists $v\in {\mathbb{R}}^n\wedge v\ne 0$, s.t. $L_1(v)\ne L_2(v)$.
$R_1(t):=f(x_0+tv)-f(x_0)-L_1(tv)$.
$R_2(t):=f(x_0+tv)-f(x_0)-L_2(tv)$.
$t(L_1(v)+L_2(v))=L_1(tv)-L_2(tv)=R_2(t)-R_1(t)$.
$\frac{||(L_1(v)-L_2(v))||}{||v||}=\frac{||t(L_1(v)-L_2(v))||}{||tv||}=\frac{||R_2(t)-R_1(t)||}{||tv||}\le \frac{||R_2(t)||}{||tv||}+\frac{||R_1(t)||}{||tv||}$.
so $L_1(v)-L_2(v)=0$, contradiction.

Properties

1. Let $F$ be a field, $⟨F,F^n⟩=:F^n$, $f:F^n\to F^m$ be a linear mapping. Then $\exists L\in F^{m\times n}$, s.t. $f(x)=Lx$. Furthermore, $L$ is unique.
2. Suppose $f$ and $g$ are defined on $[a,b]$ and are differentiable at a point $x\in [a,b]$. Then $f+g$, $f\cdot g$, and $\frac{f}{g}$ are differentiable at $x$, and
   1. $(f+g{)}^{\prime }(x)=f^{\prime }(x)+g^{\prime }(x)$
   2. $(fg{)}^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$
   3. ${\left(\frac{f}{g}\right)}^{\prime }(x)=\frac{g(x)f^{\prime }(x)-g^{\prime }(x)f(x)}{g^2(x)}$
3. Let $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ be a function, and $f$ is differentiable at $x_0$ on $X$. $g:{\mathbb{R}}^m\to {\mathbb{R}}^k$ be a function, and $g$ is differentiable at $f(x_0)$ on $f(X)$. Then $g\circ f$ is differentiable at $x_0$ on $X$, and $(g\circ f{)}^{\prime }(x_0)=g^{\prime }(f(x_0))\circ f^{\prime }(x_0)$.

Proofs

1.

• Let $F$ be a field, $⟨F,F^n⟩=:F^n$, $f:F^n\to F^m$ be a linear mapping. Then $\exists L\in F^{m\times n}$, s.t. $f(x)=Lx$. Furthermore, $L$ is unique.

Let $(e_i{)}_{i=1}^n$ be the standard basis of $F^n$, then $f(e_i)=L_i$, $L_i\in F^m$. $L:=(L_1,L_2,\cdots ,L_n{)}^T$.

$x$ can be written as $x={\sum }_{i=1}^n{x}_i{e}_i$, then $f(x)={\sum }_{i=1}^n{x}_{i}f(e_i)={\sum }_{i=1}^n{x}_i{L}_i=Lx$

uniqueness is obvious.

2.

• Let $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ be a function, and $f$ is differentiable at $x_0$ on $X$. $g:{\mathbb{R}}^m\to {\mathbb{R}}^k$ be a function, and $g$ is differentiable at $f(x_0)$ on $f(X)$. Then $g\circ f$ is differentiable at $x_0$ on $X$, and $(g\circ f{)}^{\prime }(x_0)=g^{\prime }(f(x_0))\circ f^{\prime }(x_0)$.