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Partial Derivatives: Derivatives of Multivariable Vector Functions 📂Vector Analysis

Partial Derivatives: Derivatives of Multivariable Vector Functions

Buildup[^1]

Recall the definition of the derivative of a univariate function.

limh0f(x+h)f(x)h=f(x) \lim \limits_{h\to 0} \dfrac{f(x+h) - f(x)}{h} = f^{\prime}(x)

By approximating the numerator on the left-hand side as a linear function of hh, we get the following.

f(x+h)f(x)=ah+r(h) \begin{equation} f(x+h) - f(x) = a h + r(h) \label{1} \end{equation}

Let’s call r(h)r(h) the remainder, satisfying the condition below.

limh0r(h)h=0 \lim \limits_{h \to 0} \dfrac{r(h)}{h}=0

Then, dividing both sides of (1)\eqref{1} by hh and taking the limit limh0\lim_{h\to 0}, we get the following.

limh0f(x+h)f(x)h=limh0ah+r(h)h=a+limh0r(h)h=a \lim \limits_{h\to 0} \dfrac{f(x+h) - f(x)}{h} = \lim \limits_{h\to 0} \dfrac{ah+ r(h)}{h} = a + \lim \limits_{h\to 0} \dfrac{r(h)}{h} = a

Here, aa was the coefficient of the first-order term in the linear approximation of hh. In this sense, aa is referred to as the derivative “coefficient” of ff at xx. By slightly transforming the above equation, we can see that the derivative coefficient of ff at xx satisfies the equation for aa.

limh0f(x+h)f(x)ahh=limh0r(h)h=0 \lim \limits_{h\to 0} \dfrac{f(x+h) - f(x) - ah}{h} = \lim \limits_{h\to 0} \dfrac{r(h)}{h} = 0

This forms the basis for defining the derivative of a multivariable vector function.

Definition

Let’s denote ERnE\subset \mathbb{R}^{n} as an open set, and xE\mathbf{x}\in E accordingly. For f:ERm\mathbf{f} : E \to \mathbb{R}^{m}, if there exists a linear transformation AL(Rn,Rm)A\in L(\mathbb{R}^{n}, \mathbb{R}^{m}) for hRn\mathbf{h} \in \mathbb{R}^{n} that satisfies the following, then ff is differentiable at x\mathbf{x}. Furthermore, AA is called the total derivative or simply the derivative of ff and is denoted by f(x)\mathbf{f}^{\prime}(\mathbf{x}).

limh0f(x+h)f(x)A(h)h=0 \begin{equation} \lim \limits_{|\mathbf{h}| \to 0} \dfrac{| \mathbf{f} ( \mathbf{x} + \mathbf{h}) - \mathbf{f} (\mathbf{x}) - A( \mathbf{h} )|}{|\mathbf{h}|} = 0 \label{2} \end{equation}

If f\mathbf{f} is differentiable at all points in EE, then f\mathbf{f} is said to be differentiable in EE.

Explanation

The term “total” implies entirety, contrasting with partial derivatives. It’s not the total ˇ\check{} function, but the total ˇ\check{} derivative.

It is important to note that f(x)\mathbf{f}^{\prime}(\mathbf{x}) is not a function value but a linear transformation satisfying f(x):ERnRm\mathbf{f}^{\prime}(\mathbf{x}) : E \subset \R^{n} \to \R^{m}. Therefore, f(x)=A\mathbf{f}^{\prime}(\mathbf{x}) = A can be represented as a matrix as follows.

f(x)=A=[a11a12a1na21a22a2nam1am2amn] \mathbf{f}^{\prime}(\mathbf{x}) = A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}

Then, the total derivative of f\mathbf{f}, f\mathbf{f}^{\prime}, can be seen as a function mapping a certain matrix AA every time xERn\mathbf{x} \in E \subset \R^{n} is provided. This matrix can be easily obtained from the partial derivatives of f\mathbf{f} and is also known as the Jacobian matrix.

f(x)=[(D1f1)(x)(D2f1)(x)(Dnf1)(x)(D1f2)(x)(D2f2)(x)(Dnf2)(x)(D1fm)(x)(D2fm)(x)(Dnfm)(x)] \mathbf{f}^{\prime}(\mathbf{x}) = \begin{bmatrix} (D_{1}f_{1}) (\mathbf{x}) & (D_{2}f_{1}) (\mathbf{x}) & \cdots & (D_{n}f_{1}) (\mathbf{x}) \\ (D_{1}f_{2}) (\mathbf{x}) & (D_{2}f_{2}) (\mathbf{x}) & \cdots & (D_{n}f_{2}) (\mathbf{x}) \\ \vdots & \vdots & \ddots & \vdots \\ (D_{1}f_{m}) (\mathbf{x}) & (D_{2}f_{m}) (\mathbf{x}) & \cdots & (D_{n}f_{m}) (\mathbf{x}) \end{bmatrix}

The total derivative is the ultimate generalization of differentiation for functions defined on finite dimensions, extending the domain and range of f\mathbf{f} to Banach spaces as the Fréchet derivative. The properties that held for univariate functions naturally hold as well.

  • Uniqueness
  • Chain Rule

Theorem

Uniqueness

Let E,x,fE, \mathbf{x}, \mathbf{f} be as defined in the Definition. If A1,A2A_{1}, A_{2} satisfies (2)\eqref{2}, then the two linear transformations are equal.

A1=A2 A_{1} = A_{2}

Proof

Let’s denote B=A1A2B = A_{1} - A_{2}. Then, by the triangle inequality, the following holds.

B(h)=A1(h)A2(h)=A1(h)f(x+h)f(x)+f(x+h)+f(x)A2(h)f(x+h)+f(x)A1(h)+f(x+h)+f(x)A2(h) \begin{align*} | B( \mathbf{h} ) | &= \left| A_{1}(\mathbf{h}) - A_{2}(\mathbf{h}) \right| \\ &= | A_{1}(\mathbf{h}) - \mathbf{f} (\mathbf{x} + \mathbf{h}) - \mathbf{f} (\mathbf{x}) + \mathbf{f} (\mathbf{x} + \mathbf{h}) + \mathbf{f} (\mathbf{x}) - A_{2}(\mathbf{h}) | \\ &\le | \mathbf{f} (\mathbf{x} + \mathbf{h}) + \mathbf{f} (\mathbf{x}) - A_{1}(\mathbf{h}) | + | \mathbf{f} (\mathbf{x} + \mathbf{h}) + \mathbf{f} (\mathbf{x}) - A_{2}(\mathbf{h}) | \end{align*}

Then, for a fixed h0\mathbf{h} \ne \mathbf{0}, the equation below holds.

limt0B(th)thlimt0f(x+th)+f(x)A1(th)th+limt0f(x+th)+f(x)A2(th)th=0 \lim _{t \to 0} \dfrac{ | B( t\mathbf{h} ) |}{| t\mathbf{h} |} \le \lim _{t \to 0}\dfrac{ | \mathbf{f} (\mathbf{x} + t\mathbf{h}) + \mathbf{f} (\mathbf{x}) - A_{1}(t\mathbf{h}) |}{| t\mathbf{h} |} + \lim _{t \to 0}\dfrac{| \mathbf{f} (\mathbf{x} + t\mathbf{h}) + \mathbf{f} (\mathbf{x}) - A_{2}(t\mathbf{h}) |}{| t\mathbf{h} |}=0

However, since BB is a linear transformation, the left-hand side is independent of tt.

limt0tB(h)th=limt0B(h)h=B(h)h0 \lim _{t \to 0} \dfrac{ | tB( \mathbf{h} ) |}{| t\mathbf{h} |} = \lim _{t \to 0} \dfrac{ | B( \mathbf{h} ) |}{| \mathbf{h} |} = \dfrac{ | B( \mathbf{h} ) |}{| \mathbf{h} |} \le 0

Since h0\mathbf{h} \ne \mathbf{0}, for the above equation to hold, it must be B=0B=0. Thus, we obtain the following.

B=A1A2=0    A1=A2 B=A_{1}-A_{2}=0 \implies A_{1} = A_{2}

Chain Rule

As defined, let’s consider ERnE \subset \R^{n} as an open set and f:ERm\mathbf{f} : E \to \R^{m} as a function differentiable in x0E\mathbf{x}_{0} \in E. Let g:f(E)Rk\mathbf{g} : \mathbf{f}(E) \to \R^{k} be a function differentiable in f(x0)f(E)\mathbf{f}(\mathbf{x}_{0}) \in \mathbf{f}(E). Also, let’s consider F:ERk\mathbf{F} : E \to \R^{k} as the composition of f\mathbf{f} and g\mathbf{g}.

F(x)=g(f(x)) \mathbf{F} (\mathbf{x}) = \mathbf{g} \left( \mathbf{f}(\mathbf{x}) \right)

Then, F\mathbf{F} is differentiable in x0\mathbf{x}_{0}, and the total derivative is as follows.

F(x0)=g(f(x0))f(x0) \mathbf{F}^{\prime} (\mathbf{x}_{0}) = \mathbf{g}^{\prime} \left( \mathbf{f}(\mathbf{x}_{0}) \right) \mathbf{f}^{\prime} (\mathbf{x}_{0})

Proof

Generalized proof for normed spaces