logo

Derivative of a Vector-Valued Function 📂Vector Analysis

Derivative of a Vector-Valued Function

Definition1

For the vector function r:IRR3\mathbf{r} : I \subset \mathbb{R} \to \mathbb{R}^{3}, if the following limit exists, then we say that r\mathbf{r} is differentiable at tt and its value is called the derivative of r\mathbf{r} at tt.

drdt=r(t):=limh0r(t+h)r(t)h \dfrac{d \mathbf{r}}{d t} = \mathbf{r}^{\prime}(t) := \lim_{h \to 0} \dfrac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h}

If for all tIt \in I there exists r(t)\mathbf{r}^{\prime}(t), then we say that r\mathbf{r} is differentiable at II. When r\mathbf{r} is differentiable at II, the r\mathbf{r}^{\prime} defined on II is called the derivative of r\mathbf{r}.

Explanation

This is a direct extension of the definition of the derivative for the scalar function f:RRf : \mathbb{R} \to \mathbb{R}.

f(a):=limh0f(a+h)f(a)h f^{\prime} (a) := \lim_{h \to 0} {{ f (a + h ) - f(a) } \over { h }}

In the definition, even if it is Rn\mathbb{R}^{n} instead of R3\mathbb{R}^{3}, it is defined in the same way. By the theorem below, the derivative of mm order is as follows.

r(m)(t)=(f(m)(t),g(m)(t),h(m)(t)) \mathbf{r}^{(m)}(t) = \left( f^{(m)}(t), g^{(m)}(t), h^{(m)}(t) \right)

Theorem

For a differentiable function fi:RRf_{i} : \mathbb{R} \to \mathbb{R}, if r(t)=(f1(t),,fn(t))\mathbf{r}(t) = \left( f_{1}(t), \dots, f_{n}(t) \right), then

r(t)=(f1(t),,fn(t)) \mathbf{r}^{\prime}(t) = \left( f_{1}^{\prime}(t), \dots, f_{n}^{\prime}(t) \right)

Proof

It can be shown by simple calculation. By the definition of limit,

r(t)=limh0r(t+h)r(t)h=limh0(f1(t+h),,fn(t+h))(f(t),,fn(t))h=limh0(f1(t+h)f1(t)h,,fn(t+h)fn(t)h)=(limh0f1(t+h)f1(t)h,,limh0fn(t+h)fn(t)h)=(f1(t),,fn(t)) \begin{align*} \mathbf{r}^{\prime}(t) &= \lim_{h \to 0} \dfrac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h} \\ &= \lim_{h \to 0} \dfrac{(f_{1}(t+h), \dots, f_{n}(t+h)) - (f(t), \dots, f_{n}(t))}{h} \\ &= \lim_{h \to 0} \left( \dfrac{f_{1}(t+h) - f_{1}(t)}{h}, \dots, \dfrac{f_{n}(t+h) - f_{n}(t)}{h} \right) \\ &= \left( \lim_{h \to 0}\dfrac{f_{1}(t+h) - f_{1}(t)}{h}, \dots, \lim_{h \to 0}\dfrac{f_{n}(t+h) - f_{n}(t)}{h} \right) \\ &= \left( f_{1}^{\prime}(t), \dots, f_{n}^{\prime}(t) \right) \end{align*}

Properties

For two vector functions u,v:RRn\mathbf{u}, \mathbf{v} : \mathbb{R} \to \mathbb{R}^{n} and scalar function f:RRf : \mathbb{R} \to \mathbb{R}, and constant cRc \in \mathbb{R}, the following hold.

1. Linearity: ddt[u(t)±v(t)]=u(t)±v(t)\dfrac{d}{dt} [\mathbf{u}(t) \pm \mathbf{v}(t)] = \mathbf{u}^{\prime}(t) \pm \mathbf{v}^{\prime}(t)

2. Linearity: ddt[cu(t)]=cu(t)\dfrac{d}{dt} [c \mathbf{u}(t)] = c \mathbf{u}^{\prime}(t)

3. Product Rule: ddt[f(t)u(t)]=f(t)u(t)+f(t)u(t)\dfrac{d}{dt} [f(t) \mathbf{u}(t)] = f^{\prime}(t) \mathbf{u}(t) + f(t) \mathbf{u}^{\prime}(t)

4. Differentiation of the Inner Product: ddt[u(t)v(t)]=u(t)v(t)+u(t)v(t)\dfrac{d}{dt} [\mathbf{u}(t) \cdot \mathbf{v}(t)] = \mathbf{u}^{\prime}(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}^{\prime}(t)

5. Differentiation of the Cross Product: ddt[u(t)×v(t)]=u(t)×v(t)+u(t)×v(t)\dfrac{d}{dt} [\mathbf{u}(t) \times \mathbf{v}(t)] = \mathbf{u}^{\prime}(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}^{\prime}(t)

6. Chain Rule: ddt[u(f(t))]=u(f(t))f(t)\dfrac{d}{dt} [\mathbf{u}(f(t))] = \mathbf{u}^{\prime}(f(t)) f^{\prime}(t)


  1. James Stewart, Daniel Clegg, and Saleem Watson, Calculus (early transcendentals, 9E), p898-899 ↩︎