# partial differentiation

• Nov 9th 2009, 06:08 AM
pepsi
partial differentiation
hello,

i am asked to prove that if z= [f(y/x)]/x then

...pls read [z_x] as the partial of z w.r.t.x...

x[z_x] + y[z_y] + z = 0 (*)

i have tried totally differentiating

dz = {-[1/x^2][f(y/x)] + [1/x][f_x] }dx + [1/x][f_y]dy

but then i get from the LHS of (*) that [1/x][f_x] }dx + [1/x][f_y]dy = ?

I am given clues for two methods that i may follow:

1) set y/x =v

and

2) to use the fact that for a homogenous function g(x,y) of nth order it is true that x[g_x] + y[g_y] = 0

i am stuck
• Nov 9th 2009, 06:56 AM
tonio
Quote:

Originally Posted by pepsi
hello,

i am asked to prove that if z= [f(y/x)]/x then

...pls read [z_x] as the partial of z w.r.t.x...

x[z_x] + y[z_y] + z = 0 (*)

i have tried totally differentiating

dz = {-[1/x^2][f(y/x)] + [1/x][f_x] }dx + [1/x][f_y]dy

but then i get from the LHS of (*) that [1/x][f_x] }dx + [1/x][f_y]dy = ?

I am given clues for two methods that i may follow:

1) set y/x =v

and

2) to use the fact that for a homogenous function g(x,y) of nth order it is true that x[g_x] + y[g_y] = 0

i am stuck

The first hint looks interesting: $\displaystyle v(x,y)=\frac{y}{x}\Longrightarrow z=\frac{f(v(x,y))}{x}\Longrightarrow z_x=\frac{x\,\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}-f(v)}{x^2}$ $\displaystyle =\frac{-\frac{y}{x}\,\frac{\partial f}{\partial v}-f(v)}{x^2}\,,\,\,z_y=\frac{1}{x}\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=\frac{1}{x^2}\frac{\partial f}{\partial v}$

Well, now just verify that indeed $\displaystyle xz_x+yz_y+z=0$

Tonio