91Ó°ÊÓ

A function $f\left(x,y,z\right)$is called homogeneous of degree n if $f\left(tx,ty,tz\right)=t^{n}f\left(x,y,z\right)$.

Differentiate $tx=u$with respect to $t$as:

$$\begin{gathered} \frac{d}{dt}\left(tx\right)=\frac{d}{dt}\left(u\right) \\ \frac{dt}{dt}\left(x\right)=\frac{du}{dt} \\ x=\frac{du}{dt} \end{gathered}$$

Thus,$x=\frac{du}{dt}\left(1\right)$

Now differentiate$ty=v$ with respect to $t$as:

$$\begin{gathered} \frac{d}{dt}\left(ty\right)=\frac{d}{dt}\left(v\right) \\ \frac{dt}{dt}\left(y\right)=\frac{dv}{dt} \\ y=\frac{dv}{dt} \end{gathered}$$

Thus,$y=\frac{dv}{dt}\left(2\right)$

Now differentiate$tz=w$ with respect to $t$as:

$$\begin{gathered} \frac{d}{dt}\left(tz\right)=\frac{d}{dt}\left(w\right) \\ \frac{dt}{dt}\left(z\right)=\frac{dw}{dt} \\ z=\frac{dw}{dt} \end{gathered}$$

Thus,$z=\frac{dw}{dt}\left(3\right)$

Differentiate $f\left(u,v,w\right)=t^{n}f\left(x,y,z\right)$as:

$$\begin{gathered} \frac{d}{dt}f\left(u,v,w\right)=\frac{d}{dt}\left[t^{n}f\left(x,y,z\right)\right] \\ \frac{∂f}{∂u}·\frac{du}{dt}+\frac{∂f}{∂v}·\frac{dv}{dt}+\frac{∂f}{∂w}·\frac{dw}{dt}=n{t}^{n-1}f\left(x,y,z\right) \\ \frac{∂f}{∂u}·\left(x\right)+\frac{∂f}{∂v}·\left(y\right)+\frac{∂f}{∂w}·\left(z\right)=n{t}^{n-1}f\left(x,y,z\right) \end{gathered}$$

Now from the equation , , and as:

$x\frac{∂f}{∂u}+y\frac{∂f}{∂v}+z\frac{∂f}{∂w}=n{t}^{n-1}f\left(x,y,z\right)$

Simplifying the expression as:

$x\frac{∂f}{∂\left(tx\right)}+y\frac{∂f}{∂\left(ty\right)}+z\frac{∂f}{∂\left(tz\right)}=n{t}^{n-1}f\left(x,y,z\right)$

Since,$tx=u$ , $ty=v$, and$tz=w$ as:

$$\begin{gathered} x\frac{∂f}{∂\left(1x\right)}+y\frac{∂f}{∂\left(1y\right)}+z\frac{∂f}{∂\left(1z\right)}=n{\left(1\right)}^{n-1}f\left(x,y,z\right) \\ x\frac{∂f}{∂x}+y\frac{∂f}{∂y}+z\frac{∂f}{∂z}=nf\left(x,y,z\right) \end{gathered}$$

Therefore, if is a homogeneous function then$x\frac{∂f}{∂x}+y\frac{∂f}{∂y}+z\frac{∂f}{∂z}=nf$ .