1 First partial derivatives
1.1 The $x$ partial derivative
For a function of a single variable, $y=f\left(x\right)$ , changing the independent variable $x$ leads to a corresponding change in the dependent variable $y$ . The rate of change of $y$ with respect to $x$ is given by the derivative , written $\frac{df}{dx}$ . A similar situation occurs with functions of more than one variable. For clarity we shall concentrate on functions of just two variables.
In the relation $z=f\left(x,y\right)$ the independent variables are $x$ and $y$ and the dependent variable $z$ . We have seen in Section 18.1 that as $x$ and $y$ vary the $z$ value traces out a surface. Now both of the variables $x$ and $y$ may change simultaneously inducing a change in $z$ . However, rather than consider this general situation, to begin with we shall hold one of the independent variables fixed . This is equivalent to moving along a curve obtained by intersecting the surface by one of the coordinate planes.
Consider $f\left(x,y\right)={x}^{3}+2{x}^{2}y+{y}^{2}+2x+1.$
Suppose we keep $y$ constant and vary $x$ ; then what is the rate of change of the function $f$ ?
Suppose we hold $y$ at the value 3 then
$\phantom{\rule{2em}{0ex}}f\left(x,3\right)={x}^{3}+6{x}^{2}+9+2x+1={x}^{3}+6{x}^{2}+2x+10$
In effect, we now have a function of $x$ only. If we differentiate it with respect to $x$ we obtain the expression:
$\phantom{\rule{2em}{0ex}}3{x}^{2}+12x+2.$
We say that $f$ has been partially differentiated with respect to $x$ . We denote the partial derivative of $f$ with respect to $x$ by $\frac{\partial f}{\partial x}$ (to be read as ‘partial dee $f$ by dee $x$ ’ ). In this example, when $y=3$ :
$\phantom{\rule{2em}{0ex}}\frac{\partial f}{\partial x}=3{x}^{2}+12x+2.$
In the same way if $y$ is held at the value 4 then $f\left(x,4\right)={x}^{3}+8{x}^{2}+16+2x+1={x}^{3}+8{x}^{2}+2x+17$ and so, for this value of $y$
$\phantom{\rule{2em}{0ex}}\frac{\partial f}{\partial x}=3{x}^{2}+16x+2.$
Now if we return to the original formulation
$\phantom{\rule{2em}{0ex}}f\left(x,y\right)={x}^{3}+2{x}^{2}y+{y}^{2}+2x+1$
and treat $y$ as a constant then the process of partial differentiation with respect to $x$ gives
$$\begin{array}{rcll}\frac{\partial f}{\partial x}& =& 3{x}^{2}+4xy+0+2+0\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& 3{x}^{2}+4xy+2.& \text{}\end{array}$$Key Point 1
The Partial Derivative of $f$ with respect to $x$
For a function of two variables $z=f\left(x,y\right)$ the partial derivative of $f$ with respect to $x$ is denoted by $\frac{\partial f}{\partial x}$ and is obtained by differentiating $f\left(x,y\right)$ with respect to $x$ in the usual way but treating the $y$ variable as if it were a constant.
Alternative notations for $\frac{\partial f}{\partial x}$ are ${f}_{x}\left(x,y\right)$ or ${f}_{x}$ or $\frac{\partial z}{\partial x}$ .
Example 2
Find $\frac{\partial f}{\partial x}$ for
 $f\left(x,y\right)={x}^{3}+x+{y}^{2}+y$ ,
 $f\left(x,y\right)={x}^{2}y+x{y}^{3}.$
Solution
 $\frac{\partial f}{\partial x}=3{x}^{2}+1+0+0=3{x}^{2}+1$
 $\frac{\partial f}{\partial x}=2x\times y+1\times {y}^{3}=2xy+{y}^{3}$
1.2 The $y$ partial derivative
For functions of two variables $f\left(x,y\right)$ the $x$ and $y$ variables are on the same footing, so what we have done for the $x$ variable we can do for the $y$ variable. We can thus imagine keeping the $x$ variable fixed and determining the rate of change of $f$ as $y$ changes. This rate of change is denoted by $\frac{\partial f}{\partial y}$ .
Key Point 2
The Partial Derivative of $f$ with respect to $y$
For a function of two variables $z=f\left(x,y\right)$ the partial derivative of $f$ with respect to $y$ is denoted by $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}$ and is obtained by differentiating $f\left(x,y\right)$ with respect to $y$ in the usual way but treating the $x$ variable as if it were a constant.
Alternative notations for $\frac{\partial f}{\partial y}$ are ${f}_{y}\left(x,y\right)$ or ${f}_{y}$ or $\frac{\partial z}{\partial y}$ .
Returning to $f\left(x,y\right)={x}^{3}+2{x}^{2}y+{y}^{2}+2x+1$ once again, we therefore obtain:
$\phantom{\rule{2em}{0ex}}\frac{\partial f}{\partial y}=0+2{x}^{2}\times 1+2y+0+0=2{x}^{2}+2y.$
Example 3
Find $\frac{\partial f}{\partial y}$ for
 $f\left(x,y\right)={x}^{3}+x+{y}^{2}+y$
 $f\left(x,y\right)={x}^{2}y+x{y}^{3}$
Solution
 $\frac{\partial f}{\partial y}=0+0+2y+1=2y+1$
 $\frac{\partial f}{\partial y}={x}^{2}\times 1+x\times 3{y}^{2}={x}^{2}+3x{y}^{2}$
We can calculate the partial derivative of $f$ with respect to $x$ and the value of $\frac{\partial f}{\partial x}$ at a specific point e.g. $x=1,\phantom{\rule{1em}{0ex}}y=2$ .
Example 4
Find ${f}_{x}\left(1,2\right)$ and ${f}_{y}\left(3,2\right)$ for $f\left(x,y\right)={x}^{2}+{y}^{3}+2xy$ .
[Remember ${f}_{x}$ means $\frac{\partial f}{\partial x}$ and ${f}_{y}$ means $\frac{\partial f}{\partial y}$ .]
Solution
${f}_{x}\left(x,y\right)=2x+2y$ , so ${f}_{x}\left(1,2\right)=24=2$ ; ${f}_{y}\left(x,y\right)=3{y}^{2}+2x,$ so ${f}_{y}\left(3,2\right)=126=6$
Task!
Given $f\left(x,y\right)=3{x}^{2}+2{y}^{2}+x{y}^{3}$ find ${f}_{x}\left(1,2\right)$ and ${f}_{y}\left(1,1\right)$ .
First find expressions for $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ :
$\frac{\partial f}{\partial x}=6x+{y}^{3},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=4y+3x{y}^{2}$ Now calculate ${f}_{x}\left(1,2\right)$ and ${f}_{y}\left(1,1\right)$ :
${f}_{x}\left(1,2\right)=6\times 1+{\left(2\right)}^{3}=2$ ; ${f}_{y}\left(1,1\right)=4\times \left(1\right)+3\left(1\right)\times 1=7$
1.3 Functions of several variables
As we have seen, a function of two variables $f\left(x,y\right)$ has two partial derivatives, $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ . In an exactly analogous way a function of three variables $f\left(x,y,u\right)$ has three partial derivatives $\frac{\partial f}{\partial x}$ , $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial u}$ , and so on for functions of more than three variables. Each partial derivative is obtained in the same way as stated in Key Point 3:
Key Point 3
The Partial Derivatives of $f\left(x,y,u,v,w,\dots \phantom{\rule{0.3em}{0ex}}\right)$
For a function of several variables $z=f\left(x,y,u,v,w,\dots \phantom{\rule{0.3em}{0ex}}\right)$ the partial derivative of $f$ with respect to $v$ (say) is denoted by $\frac{\partial f}{\partial v}$ and is obtained by differentiating $f\left(x,y,u,v,w,\dots \phantom{\rule{0.3em}{0ex}}\right)$ with respect to $v$ in the usual way but treating all the other variables as if they were constants.
Alternative notations for $\frac{\partial f}{\partial v}$ when $z=f\left(x,y,u,v,w,\dots \phantom{\rule{0.3em}{0ex}}\right)$ are ${f}_{v}\left(x,y,u,v,w\dots \phantom{\rule{0.3em}{0ex}}\right)$ and ${f}_{v}$ and $\frac{\partial z}{\partial v}$ .
Task!
Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial u}$ for $f\left(x,y,u,v\right)={x}^{2}+x{y}^{2}+{y}^{2}{u}^{3}7u{v}^{4}$
$\frac{\partial f}{\partial x}=2x+{y}^{2}+0+0=2x+{y}^{2};\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial u}=0+0+{y}^{2}\times 3{u}^{2}7{v}^{4}=3{y}^{2}{u}^{2}7{v}^{4}$ .
Task!
The pressure, $P$ , for one mole of an ideal gas is related to its absolute temperature, $T$ , and specific volume, $v$ , by the equation
$\phantom{\rule{2em}{0ex}}Pv=RT$
where $R$ is the gas constant.
Obtain simple expressions for

the coefficient of thermal expansion,
$\alpha $
,
defined by:
$\phantom{\rule{2em}{0ex}}\alpha =\frac{1}{v}{\left(\frac{\partial v}{\partial T}\right)}_{P}$

the isothermal compressibility,
${\kappa}_{T}$
,
defined by:
$\phantom{\rule{2em}{0ex}}{\kappa}_{T}=\frac{1}{v}{\left(\frac{\partial v}{\partial P}\right)}_{T}$
$\phantom{\rule{2em}{0ex}}v=\frac{RT}{P}\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{\left(\frac{\partial v}{\partial T}\right)}_{P}=\frac{R}{P}$
so $\alpha =\frac{1}{v}{\left(\frac{\partial v}{\partial T}\right)}_{P}=\frac{R}{Pv}=\frac{1}{T}$
$\phantom{\rule{2em}{0ex}}v=\frac{RT}{P}\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{\left(\frac{\partial v}{\partial P}\right)}_{T}=\frac{RT}{{P}^{2}}$
so ${\kappa}_{T}=\frac{1}{v}{\left(\frac{\partial v}{\partial P}\right)}_{T}=\frac{RT}{v{P}^{2}}=\frac{1}{P}$
Exercises

For the following functions find
$\frac{\partial f}{\partial x}$
and
$\frac{\partial f}{\partial y}$
 $f\left(x,y\right)=x+2y+3$
 $f\left(x,y\right)={x}^{2}+{y}^{2}$
 $f\left(x,y\right)={x}^{3}+xy+{y}^{3}$
 $f\left(x,y\right)={x}^{4}+x{y}^{3}+2{x}^{3}{y}^{2}$
 $f\left(x,y,z\right)=xy+yz$
 For the functions of Exercise 1 (a) to (d) find ${f}_{x}\left(1,1\right),\phantom{\rule{1em}{0ex}}{f}_{x}\left(1,1\right),\phantom{\rule{1em}{0ex}}{f}_{y}\left(1,2\right),\phantom{\rule{1em}{0ex}}{f}_{y}\left(2,1\right)$ .

 $\frac{\partial f}{\partial x}=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=2$
 $\frac{\partial f}{\partial x}=2x,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=2y$
 $\frac{\partial f}{\partial x}=3{x}^{2}+y,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=x+3{y}^{2}$
 $\frac{\partial f}{\partial x}=4{x}^{3}+{y}^{3}+6{x}^{2}{y}^{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=3x{y}^{2}+4{x}^{3}y$
 $\frac{\partial f}{\partial x}=y,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=x+z$
$$  ${f}_{x}\left(1,1\right)$  ${f}_{x}\left(1,1\right)$  ${f}_{y}\left(1,2\right)$  ${f}_{y}\left(2,1\right)$ 
(a)  1  1  2  2 
(b)  2  $2$  4  2 
(c)  4  2  13  5 
(d)  11  1  20  38 