Answer
We verify Eq. (11) for the functions in Exercise 42:
(a) $f\left( {x,y,z} \right) = {x^2}y + xyz$
(b) $f\left( {x,y,z} \right) = 3x + 2y - 8z$
(c) $f\left( {x,y,z} \right) = \ln \left( {\frac{{xy}}{{{z^2}}}} \right)$
(d) $f\left( {x,y,z} \right) = {z^4}$
Work Step by Step
Recall from Eq. (11):
(11) ${\ \ \ }$ $x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = nf$
We verify the functions in Exercise 42:
(a) $f\left( {x,y,z} \right) = {x^2}y + xyz$
The derivatives are
$\frac{{\partial f}}{{\partial x}} = 2xy + yz$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = {x^2} + xz$, ${\ \ }$ $\frac{{\partial f}}{{\partial z}} = xy$
Using Eq. (11) we get
$x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = 2{x^2}y + xyz + {x^2}y + xyz + xyz$
$ = 3{x^2}y + 3xyz$
$ = 3f\left( {x,y,z} \right)$
From Exercise 42 we know that $f\left( {x,y,z} \right)$ is homogeneous of degree $3$.
Hence Eq. (11) is verified.
(b) $f\left( {x,y,z} \right) = 3x + 2y - 8z$
The derivatives are
$\frac{{\partial f}}{{\partial x}} = 3$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = 2$, ${\ \ }$ $\frac{{\partial f}}{{\partial z}} = - 8$
Using Eq. (11) we get
$x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = 3x + 2y - 8z$
$ = f\left( {x,y,z} \right)$
From Exercise 42 we know that $f\left( {x,y,z} \right)$ is homogeneous of degree $1$.
Hence Eq. (11) is verified.
(c) $f\left( {x,y,z} \right) = \ln \left( {\frac{{xy}}{{{z^2}}}} \right)$
The derivatives are
$\frac{{\partial f}}{{\partial x}} = \frac{{{z^2}}}{{xy}}\frac{y}{{{z^2}}} = \frac{1}{x}$,
$\frac{{\partial f}}{{\partial y}} = \frac{{{z^2}}}{{xy}}\frac{x}{{{z^2}}} = \frac{1}{y}$,
$\frac{{\partial f}}{{\partial z}} = - 2\frac{{{z^2}}}{{xy}}\frac{{xy}}{{{z^3}}} = - \frac{2}{z}$
Using Eq. (11) we get
$x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = 1 + 1 - 2 = 0$
We can write
$x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = 0\cdot f\left( {x,y,z} \right)$
From Exercise 42 we know that $f\left( {x,y,z} \right)$ is homogeneous of degree $0$.
Hence Eq. (11) is verified.
(d) $f\left( {x,y,z} \right) = {z^4}$
The derivatives are
$\frac{{\partial f}}{{\partial x}} = 0$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = 0$, $\frac{{\partial f}}{{\partial z}} = 4{z^3}$
Using Eq. (11) we get
$x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = 4{z^4}$
$ = 4f\left( {x,y,z} \right)$
From Exercise 42 we know that $f\left( {x,y,z} \right)$ is homogeneous of degree $4$.
Hence Eq. (11) is verified.
