Partial Derivatives and Integrals

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:

${\begin{aligned}\quad D_{k}f(\mathbf {c} )&=\lim _{h\to 0}{\frac {f(\mathbf {c} +h\mathbf {e} _{k})-f(\mathbf {c} )}{h}}\\&=\lim _{h\to 0}{\frac {f(c_{1},c_{2},...,c_{k}+h,...,c_{n})-f(c_{1},c_{2},...,c_{k},...c_{n})}{h}}\end{aligned}}$

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

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

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

${\begin{aligned}\quad D_{1}f(1,2,-1)&=\lim _{h\to 0}{\frac {f(1+h,2,-1)-f(1,2,-1)}{h}}\\&=\lim _{h\to 0}{\frac {3(1+h)^{2}(2)(-1)-3(1)^{2}(2)(-1)}{h}}\\&=\lim _{h\to 0}{\frac {-6(1+h)^{2}+6}{h}}\\&=\lim _{h\to 0}{\frac {-6(1+2h+h^{2})+6}{h}}\\&=\lim _{h\to 0}{\frac {-12h-6h^{2}}{h}}\\&=\lim _{h\to 0}-12-6h\\&=-12\end{aligned}}$

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:

${\begin{aligned}\quad D_{1}\mathbf {f} (c)&=\lim _{h\to 0}{\frac {\mathbf {f} (c+h)-\mathbf {f} (c)}{h}}\\&=\lim _{h\to 0}{\frac {(f_{1}(c+h),f_{2}(c+h),...,f_{m}(c+h))-(f_{1}(c),f_{2}(c),...,f_{m}(c))}{h}}\\&=\lim _{h\to 0}{\frac {(f_{1}(c+h)-f_{1}(c),f_{2}(c+h)-f_{2}(c),...,f_{m}(c+h)-f(c))}{h}}\\&=\left(\lim _{h\to 0}{\frac {f_{1}(c+h)-f_{1}(c)}{h}},\lim _{h\to 0}{\frac {f_{2}(c+h)-f_{2}(c)}{h}},...,\lim _{h\to 0}{\frac {f_{m}(c+h)-f(c)}{h}}\right)\\&=(f_{1}'(c),f_{2}'(c),...,f_{m}'(c))\end{aligned}}$

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

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

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

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

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

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

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:

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

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

${\begin{aligned}\quad D_{1}\mathbf {f} (\mathbf {c} )&=\lim _{h\to 0}{\frac {\mathbf {f} (\mathbf {c} +h(1,0))-\mathbf {f} (\mathbf {c} )}{h}}\\&=\lim _{h\to 0}{\frac {(f_{1}((c_{1},c_{2})+h(1,0)),f_{2}((c_{1},c_{2})+h(1,0)))-(f_{1}(\mathbf {c} ),f_{2}(c_{1},c_{2})))}{h}}\\&=\lim _{h\to 0}{\frac {f_{1}(c_{1}+h,c_{2}),f_{2}(c_{1}+h,c_{2}))-(f_{1}(c_{1},c_{2}),f_{2}(c_{1},c_{2}))}{h}}\\&=\lim _{h\to 0}{\frac {(f_{1}(c_{1}+h,c_{2})-f_{1}(c_{1},c_{2}),f_{2}(c_{1}+h,c_{2})-f_{2}(c_{1},c_{2}))}{h}}\\&=\lim _{h\to 0}{\frac {((c_{1}+h)^{2}+c_{2}^{2}-[c_{1}^{2}+c_{2}^{2}],2(c_{1}+h)(c_{2})-2c_{1}c_{2})}{h}}\\&=\lim _{h\to 0}{\frac {(c_{1}^{2}+2c_{1}h+h^{2})+c_{2}^{2}-c_{1}^{2}-c_{2}^{2},2c_{1}c_{2}+2c_{2}h-2c_{1}c_{2})}{h}}\\&=\lim _{h\to 0}{\frac {(2c_{1}h+h^{2},2c_{2}h)}{h}}\\&=\left(\lim _{h\to 0}{\frac {2c_{1}h+h^{2}}{h}},\lim _{h\to 0}{\frac {2c_{2}h}{h}}\right)\\&=\left(\lim _{h\to 0}[2c_{1}+h],\lim _{h\to 0}2c_{2}\right)\\&=(2c_{1},2c_{2})\end{aligned}}$

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