# Chapter 1.7 Derivatives in several variables as linear transformations

The main takeaway here is that the derivative of functions in $\R^n$ can be defined as linear transformations.

Def 1.7.3. Partial derivatives. Let $U$ be an open subset of $\R^n$ and $f:U\to \R$ a function. The partial derivative of $f$ with respect to the $i$th variable, and evaluated at $a$, is the limit

$D_if(a) = \lim_{h\to 0}\frac{1}{h}\begin{pmatrix}\begin{pmatrix}a_1\\\vdots\\a_i+h\\\vdots\\a_n\end{pmatrix} - f\begin{pmatrix}a_1\\\vdots\\a_i\\\vdots\\a_n\end{pmatrix}\end{pmatrix},$

if the limit exists.

## The Jacobian Matrix

Def 1.7.7. The Jacobian Matrix. Let $U$ be an open subset of $\Reals^n$. The Jacobian Matrix of a function $f: U \mapsto \Reals^m$ is the $m\times n$ matrix composed of the $n$ partial derivatives of $f$ evaluated at $a$:

$\begin{bmatrix}\mathbf{Jf(a)}\end{bmatrix} = \begin{bmatrix} D_1f_1(a) & \dots & D_nf_1(a)\\ \vdots & \ddots & \vdots\\ D_1f_m(a) & \dots & D_nf_m(a) \end{bmatrix}$

Prop and Def 1.7.9. The Derivative. Let $U \subset \R^n$be an open subset and let $f: U \to \R^m$ be a mapping; let $a$ be a point in $U$. If there exists a linear transformation $L: \R^n \to \R^m$ such that

$\lim_{\vec{\mathbf{h}} \to \vec{\mathbf{0}}} \frac{1}{|\vec{\mathbf{h}}|} \mathbf{((f(a+\vec{h}) - f(a)) - (}L(\mathbf{\vec{h}})) = \vec{\mathbf{0}},$

then $f$ is differentiable at $a$, and $L$ is unique and is the derivative of $f$ at $a$, denoted $\mathbf{[Df(a)]}$.

Theorem. If $f$ is differentiable at $a$, then all partial derivatives of $f$ at $a$ exist, and the matrix representing $[Df(a)]$ is $[Jf(a)]$.

Proof of 1.7.7. We prove that the Jacobian Matrix is a unique linear transformation. Because the linear transformation $L$ is represented by the matrix whose $i$th column is $L(\vec{\mathbf{e_i}})$, we show that

$L(\vec{\mathbf{e_i}}) = \vec{D_i}f(a),$

where $\vec{D_i}f(a)$ is by definition the $i$th column of the Jacobian matrix $\begin{bmatrix}\mathbf{Jf(a)}\end{bmatrix}$. Using the definition of the derivative (Def 1.7.9), we can set $\vec{\mathbf{h}} = t\mathbf{\vec{e_i}}$ and let $t$ tend to 0.

$\lim_{t\vec{\mathbf{e_i}} \to \vec{\mathbf{0}}} \frac{1}{|t\vec{\mathbf{e_i}}|} \mathbf{((f(a+\mathnormal{t}\vec{e_i}) - f(a)) - (}L(t\mathbf{\vec{e_i}})) = \vec{\mathbf{0}},$
 Because $t\mathbf{\vec{e_i}} = t \mathbf{\vec{e_i}}$, and $\mathbf{\vec{e_i}} =1$, $t\mathbf{\vec{e_i}} = t$. Because the value of the limit is $\vec{\mathbf{0}}$ when $t$ is positive or negative, we replace $t$ with $t$.
$\lim_{t\vec{\mathbf{e_i}} \to \vec{\mathbf{0}}} \frac{1}{t} \mathbf{((f(a+\mathnormal{t}\vec{e_i}) - f(a)) - (}L(t\mathbf{\vec{e_i}})) = \vec{\mathbf{0}},$

Using the linearity of derivatives, we get that $L(t\vec{\mathbf{e_i}}) = tL(\vec{\mathbf{e_i}})$. Therefore, we can rewrite the equation as

$\lim_{t\vec{\mathbf{e_i}} \to \vec{\mathbf{0}}} \frac{\mathbf{((f(a+\mathnormal{t}\vec{e_i}) - f(a))}}{t}-L(\mathbf{\vec{e_i}}) = \vec{\mathbf{0}},$

The first term is the partial derivative of $f(a)$. Therefore, $L(\vec{\mathbf{e_i}}) = \vec{D_i}f(a)$. In addition, an element in $\R^n$ goes to $0$ if and only if its length goes to $0$.

Example:

The Jacobian of the mapping

$f\dbinom{x}{y} = \dbinom{xy}{x^2 - y^2} \text{ is } \begin{bmatrix}Jf\dbinom{x}{y}\end{bmatrix} = \begin{bmatrix}y&x\\2x&-2y\end{bmatrix}$

## Directional Derivatives

Directional derivatives are a generalization of partial derivatives. The partial derivative $\vec{D_i}f(a)$ describes how $f(x)$ varies as the variable $x$ moves from $a$ in the direction of the standard basis vector $\vec{\bold{e_i}}$.

The directional derivative describes how $f(x)$ varies when the variable moves in any direction $\mathbf{\vec{v}}$, i.e. when the variable $x$ moves at a constant speed from $a$ to $a+ \vec{\mathbf{v}}$ is

$\lim_{h\to0}\frac{f(a+h\vec{\mathbf{v}})-f(a)}{h}$

Proposition 1.7.14. If $U \in \R^n$ is open, and $f: U \to \R^n$ is differentiable at $a \in U$, then all direction derivatives of $f$ at $a$ exist, and the direction derivative in the direction $\vec{\mathbf{v}}$ is given by the formula

$\lim_{h\to0}\frac{f(a+h\vec{\mathbf{v}})-f(a)}{h} = [Df(a)]\vec{\mathbf{v}}$

Example: Compute the derivative in the direction $\vec{\mathbf{v}} = \begin{bmatrix}1\2\1\end{bmatrix}$ of the function $f\begin{pmatrix}x\y\z\end{pmatrix} = xy\sin z$, evaluated at $a = \begin{pmatrix}1\1\ \pi /2 \end{pmatrix}$.

First, we take its derivative.

$\begin{bmatrix}Df(x,y,z)\end{bmatrix} = \begin{bmatrix}y\sin z & x\sin z& xy \cos z\end{bmatrix}$

At $a$, this is $[1, 1, 0]$. We use Prop 1.7.14 to compute the directional derivative.

$[Df(x,y,z)]\vec{\mathbf{v}} = \begin{bmatrix}1,1,0\end{bmatrix}\begin{bmatrix}1\\2\\1\end{bmatrix} = 3$

Example: Let $f: \R^2 \to \R^2$ be the function $f\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}xy\ x^2 - y^2 \end{pmatrix}$. At the point $\begin{pmatrix}1\1\end{pmatrix}$, wat what rate is $f$ varying in the direction $\begin{bmatrix}2\1\end{bmatrix}$?

$[Df(x,y)] = \begin{bmatrix}y & x \\2x & -2y \end{bmatrix}$ $[Df(1,1)] = \begin{bmatrix}1 & 1 \\2 & -2 \end{bmatrix}$ $[Df(1,1)]\vec{\mathbf{v}} = \begin{bmatrix}1 & 1 \\2 & -2 \end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix} = \begin{bmatrix}3\\2\end{bmatrix}$

## The Jacobian matrix: not always the right approach

Although computing derivatives with the Jacobian matrix is handy, it doesn’t always work. In contrast, taking the limit as $\vec{\mathbf{h}} \to \vec{\mathbf{0}}$ always works.

Proposition 1.7.18 If $f$ is the function $f(A) = A^{-1}$, defined on the set of invertible matricies in $\text{Mat}(n,n)$, then $f$ is differentiable and

$[Df(A)]H = -A^{-1}HA^{-1}$

Proof. For $f$ to be differentiable, it must be defined on an open subset of $\text{Mat}(n,n)$. Because we know that the subset of invertible matrices is open, we can take the derivative. We show that

$\lim_{H\to [0]}\frac{1}{|H|}\underbrace{((A+H)^{-1}-A^{-1}}_{\text{increment to mapping}} - \underbrace{-A^{-1}HA^{-1}}_{\text{linear function of H}}) = [0]$

Because $H \to [0]$, we can assume that $[AH^{-1}] < 1$. Therefore, we can treat $[I A^{-1}H]$ as a sum of a geometric series.

$(A+H)^{-1} = (A(I+A^{-1}H))^{-1} = (I+A^{-1}H)^{-1}A^{-1}$ $=(I-(-A^{-1}H))A^{-1}$

We replace $I -(-A^{-1}H)$with the infinite series $I + (-A^{-1}H )+ (-A^{-1}H )^2 + \dots$.

$= (I + (-A^{-1}H )+ (-A^{-1}H )^2 + \dots)A^{-1}$ $= A^{-1}-A^{-1}HA^{-1}+((-A^{-1}H)^2+(-A^{-1}H)^3+\dots)A^{-1}$

Subtracting $A^{-1}-A^{-1}HA^{-1}$ from both sides, we get the increment and its linear approximation:

$\underbrace{((A+H)^{-1}-A^{-1})}_{f(a+h)-f(a)}-\underbrace{(-A^{-1}HA^{-1})}_{\text{linear function of increment H}}$ $=((-A^{-1}H)^2+(-A^{-1}H)^3+\dots)A^{-1}$ $=(-A^{-1}H)((-A^{-1}H)+(-A^{-1}H)^2+(-A^{-1}H)^3+\dots)A^{-1}$

The Cauchy-Schwarz Inequality gives us

$((A+H)^{-1}-A^{-1})+A^{-1}HA^{-1})\leq |A^{-1}H||(-A^{-1}H)+(-A^{-1}H)^2+(-A^{-1}H)^3+\dots||A^{-1}|$

And the triangle inequality gives us

$\leq |A^{-1}H||A^{-1}||-A^{-1}H|+|-A^{-1}H|^2+|-A^{-1}H|^3+\dots$ $\leq |A^{-1}|^2|H|\frac{|A^{-1}||H|}{1-|A^{-1}H|}\leq \frac{|A^{-1}|^3|H|^2}{1-|A^{-1}H|}$
 If $H$ is so small that $A^{-1}H < 1/2$, then
$\frac{1}{1-|A^{-1}H|} \leq 2$

We see that

$\lim_{H\to [0]}\frac{1}{|H|}{((A+H)^{-1}-A^{-1}} +A^{-1}HA^{-1}) \leq\lim_{H\to [0]} \frac{1}{|H|}2|H|^2|A^{-1}|^3= [0]$