Partial Derivatives of Functions from Rn to Rm

One of the core concepts of multivariable calculus involves the various differentiations of functions from $\mathbb{R}^n$ to $\mathbb{R}^m$. We begin by defining the concept of a partial derivative of such functions.

Definition: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$. Denote $\mathbf{e}_k = (0, 0, ..., 0, \underbrace{1}_{k^{\mathrm{th}} \: coordinate}, 0, ..., 0) \in \mathbb{R}^n$ for each $k \in \{ 1, 2, ..., n \}$, i.e., $\mathbf{e}_k$ is the unit vector in the direction of the $k^{\mathrm{th}}$ coordinate axis. Then the Partial Derivative of $\mathbf{f}$ at $\mathbf{c}$ with Respect to the $k^{\mathrm{th}}$ Variable is defined as $\displaystyle{D_k \mathbf{f} (\mathbf{c}) = \lim_{h \to 0} \frac{\mathbf{f} (\mathbf{c} + h\mathbf{e}_k) - \mathbf{f}(\mathbf{c})}{h}}$ provided that this limit exists.

Suppose that $S \subseteq \mathbb{R}^n$ is open, $\mathbf{c} \in S$, and $f : S \to \mathbb{R}$. Then the partial derivative of $f$ at $\mathbf{c}$ with respect to the $k^{\mathrm{th}}$ variable is:

(1)

\begin{align} \quad D_k f(\mathbf{c}) &= \lim_{h \to 0} \frac{f(\mathbf{c} + h\mathbf{e}_k) - f(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{f(c_1, c_2, ..., c_k + h, ..., c_n) - f(c_1, c_2, ..., c_k, ... c_n)}{h} \end{align}

For example, consider the function $f : \mathbb{R}^3 \to \mathbb{R}$ defined by:

(2)

\begin{align} \quad f(x, y, z) = 3x^2yz \end{align}

Then the partial derivative of $f$ with respect to the variable $x$ at the point $(1, 2, -1)$ is:

(3)

\begin{align} \quad D_1 f(1, 2, -1) &= \lim_{h \to 0} \frac{f(1 + h, 2, -1) - f(1, 2, -1)}{h} \\ &= \lim_{h \to 0} \frac{3(1 + h)^2(2)(-1) - 3(1)^2(2)(-1)}{h} \\ &= \lim_{h \to 0} \frac{-6(1 + h)^2 + 6}{h} \\ &= \lim_{h \to 0} \frac{-6(1 + 2h + h^2) + 6}{h} \\ &= \lim_{h \to 0} \frac{-12h - 6h^2}{h} \\ &= \lim_{h \to 0} -12 - 6h \\ &= -12 \end{align}

We can also easily calculate the partial derivatives $D_2 f(1, 2, -1)$ and $D_3(1, 2, -1)$. So the definition of a partial derivative for $\mathbf{f} : S \to \mathbb{R}^m$ is somewhat justified since the case when $m = 1$ yields the definition of the partial derivative for a multivariable real-valued function.

Furthermore, suppose that $S \subseteq \mathbb{R}$ and that $\mathbf{f} : S \to \mathbb{R}^m$. Then $\mathbf{f} = (f_1, f_2, ..., f_m)$ where $f_i : S \to \mathbb{R}$ for each $i \in \{ 1, 2, ..., m \}$ are single-variable real-valued functions. The partial derivative of $\mathbf{f}$ with respect to the first variable (the only variable, or simply just the derivative) at $\mathbf{c}$ is:

(4)

\begin{align} \quad D_1 \mathbf{f} (c) &= \lim_{h \to 0} \frac{\mathbf{f}(c + h) - \mathbf{f}(c)}{h} \\ &= \lim_{h \to 0} \frac{(f_1(c + h), f_2(c + h), ..., f_m(c + h)) - (f_1(c), f_2(c), ..., f_m(c))}{h} \\ &= \lim_{h \to 0} \frac{(f_1(c + h) - f_1(c), f_2(c + h) - f_2(c), ..., f_m(c + h) - f(c))}{h} \\ &= \left ( \lim_{h \to 0} \frac{f_1(c + h) - f_1(c)}{h}, \lim_{h \to 0} \frac{f_2(c + h) - f_2(c)}{h}, ..., \lim_{h \to 0} \frac{f_m(c + h) - f(c)}{h} \right ) \\ &= (f_1'(c), f_2'(c), ..., f_m'(c)) \end{align}

For example, consider the function $f : \mathbb{R} \to \mathbb{R}^4$ defined by:

(5)

\begin{align} \quad \mathbf{f}(t) = (t, t^2, t^3, t^4) \end{align}

Then the derivative of $\mathbf{f}$ is:

(6)

\begin{align} \quad \mathbf{f}(t) = (1, 2t, 3t^2, 4t^3) \end{align}

And the derivative of $\mathbf{f}$ at $c = 2$ is:

(7)

\begin{align} \quad \mathbf{f}(2) = (1, 4, 12, 32) \end{align}

Once again, the definition is justified since when $n = 1$ we have that the definition reduces down to the special case of differentiating a single variable vector-valued function.

Now let's look at a more complicated example of computing a partial derivative. Let $\mathbf{f} : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by:

(8)

\begin{align} \quad \mathbf{f}(x, y) = (x^2 + y^2, 2xy) \end{align}

Then the partial derivative of $\mathbf{f}$ at $\mathbf{c}$ with respect to the first variable is:

(9)

\begin{align} \quad D_1 \mathbf{f}(\mathbf{c}) &= \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c} + h(1, 0)) - \mathbf{f}(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{(f_1((c_1, c_2)+ h(1, 0)), f_2((c_1, c_2)+ h(1, 0))) - (f_1(\mathbf{c}), f_2(c_1, c_2)))}{h} \\ &= \lim_{h \to 0} \frac{f_1(c_1 + h, c_2), f_2(c_1 + h, c_2)) - (f_1(c_1, c_2), f_2(c_1, c_2))}{h} \\ &= \lim_{h \to 0} \frac{(f_1(c_1 + h, c_2) - f_1(c_1, c_2), f_2(c_1 + h, c_2) - f_2(c_1, c_2))}{h} \\ &= \lim_{h \to 0} \frac{((c_1 + h)^2 + c_2^2 - [c_1^2 + c_2^2], 2(c_1 + h)(c_2) - 2c_1c_2)}{h} \\ &= \lim_{h \to 0} \frac{(c_1^2 + 2c_1h + h^2) + c_2^2 - c_1^2 - c_2^2, 2c_1c_2 + 2c_2h - 2c_1c_2)}{h} \\ &= \lim_{h \to 0} \frac{(2c_1h + h^2, 2c_2h)}{h} \\ &= \left ( \lim_{h \to 0} \frac{2c_1h + h^2}{h}, \lim_{h \to 0} \frac{2c_2h}{h} \right ) \\ &= \left ( \lim_{h \to 0} [2c_1 + h], \lim_{h \to 0} 2c_2 \right ) \\ &= (2c_1, 2c_2) \end{align}

So the partial derivative of $\mathbf{f}$ with respect to the first variable at say $(1, 2)$ is $D_1 \mathbf{f}(1, 2) = (2, 4)$.