### 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

1. $f\left(x,y\right)={x}^{3}+x+{y}^{2}+y$ ,
2. $f\left(x,y\right)={x}^{2}y+x{y}^{3}.$
##### Solution
1. $\frac{\partial f}{\partial x}=3{x}^{2}+1+0+0=3{x}^{2}+1$
2. $\frac{\partial f}{\partial x}=2x×y+1×{y}^{3}=2xy+{y}^{3}$

#### 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:

$\phantom{\rule{2em}{0ex}}\frac{\partial f}{\partial y}=0+2{x}^{2}×1+2y+0+0=2{x}^{2}+2y.$

##### Example 3

Find $\frac{\partial f}{\partial y}$ for

1. $f\left(x,y\right)={x}^{3}+x+{y}^{2}+y$
2. $f\left(x,y\right)={x}^{2}y+x{y}^{3}$
##### Solution
1. $\frac{\partial f}{\partial y}=0+0+2y+1=2y+1$
2. $\frac{\partial f}{\partial y}={x}^{2}×1+x×3{y}^{2}={x}^{2}+3x{y}^{2}$

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 ${f}_{x}\left(1,-2\right)=2-4=-2$ ; ${f}_{y}\left(x,y\right)=3{y}^{2}+2x,$ so ${f}_{y}\left(-3,2\right)=12-6=6$

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 ${f}_{y}\left(-1,-1\right)$ :

${f}_{x}\left(1,-2\right)=6×1+{\left(-2\right)}^{3}=-2$ ; ${f}_{y}\left(-1,-1\right)=4×\left(-1\right)+3\left(-1\right)×1=-7$

#### 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 $\frac{\partial z}{\partial v}$ .

Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial u}$ for $f\left(x,y,u,v\right)={x}^{2}+x{y}^{2}+{y}^{2}{u}^{3}-7u{v}^{4}$

$\frac{\partial f}{\partial x}=2x+{y}^{2}+0+0=2x+{y}^{2};\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial u}=0+0+{y}^{2}×3{u}^{2}-7{v}^{4}=3{y}^{2}{u}^{2}-7{v}^{4}$ .

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

1. the coefficient of thermal expansion, $\alpha$ , defined by:

$\phantom{\rule{2em}{0ex}}\alpha =\frac{1}{v}{\left(\frac{\partial v}{\partial T}\right)}_{P}$

2. the isothermal compressibility, ${\kappa }_{T}$ , defined by:

$\phantom{\rule{2em}{0ex}}{\kappa }_{T}=-\frac{1}{v}{\left(\frac{\partial v}{\partial P}\right)}_{T}$

$\phantom{\rule{2em}{0ex}}v=\frac{RT}{P}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{\left(\frac{\partial v}{\partial T}\right)}_{P}=\frac{R}{P}$

so $\alpha =\frac{1}{v}{\left(\frac{\partial v}{\partial T}\right)}_{P}=\frac{R}{Pv}=\frac{1}{T}$

$\phantom{\rule{2em}{0ex}}v=\frac{RT}{P}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{\left(\frac{\partial v}{\partial P}\right)}_{T}=-\frac{RT}{{P}^{2}}$

so ${\kappa }_{T}=-\frac{1}{v}{\left(\frac{\partial v}{\partial P}\right)}_{T}=\frac{RT}{v{P}^{2}}=\frac{1}{P}$

##### Exercises
1. For the following functions find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$
1. $f\left(x,y\right)=x+2y+3$
2. $f\left(x,y\right)={x}^{2}+{y}^{2}$
3. $f\left(x,y\right)={x}^{3}+xy+{y}^{3}$
4. $f\left(x,y\right)={x}^{4}+x{y}^{3}+2{x}^{3}{y}^{2}$
5. $f\left(x,y,z\right)=xy+yz$
2. For the functions of Exercise 1 (a) to (d) find ${f}_{x}\left(1,1\right),\phantom{\rule{1em}{0ex}}{f}_{x}\left(-1,-1\right),\phantom{\rule{1em}{0ex}}{f}_{y}\left(1,2\right),\phantom{\rule{1em}{0ex}}{f}_{y}\left(2,1\right)$ .
1. $\frac{\partial f}{\partial x}=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=2$
2. $\frac{\partial f}{\partial x}=2x,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=2y$
3. $\frac{\partial f}{\partial x}=3{x}^{2}+y,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=x+3{y}^{2}$
4. $\frac{\partial f}{\partial x}=4{x}^{3}+{y}^{3}+6{x}^{2}{y}^{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=3x{y}^{2}+4{x}^{3}y$
5. $\frac{\partial f}{\partial x}=y,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=x+z$

  ${f}_{x}\left(1,1\right)$ ${f}_{x}\left(-1,-1\right)$ ${f}_{y}\left(1,2\right)$ ${f}_{y}\left(2,1\right)$ (a) 1 1 2 2 (b) 2 $-2$ 4 2 (c) 4 2 13 5 (d) 11 1 20 38