Multivariable calculus proof involving the partial derivatives of an expression

[KungPeng Zhou]

Homework Statement

Supposed f is a function of several variables that satisfies the equation f(tx, ty, tz) =t^{n}f(x, y, z),(t as any number).
Prove:
##xf_{x}+yf_{y}+zf_{z}=nf(x, y, z) ##

Relevant Equations

Partial derivative related formulas

For the first equation:
##f(tx, ty, tz)=f(u, v, w) ##, ##u=tx, v=ty, w=tz##,##k=f(u, v, w) ####
t^{n}f_{x}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}##
As the same calculation
##xf_{x}+yf_{y}+zf_{z}=[\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} +\frac{\partial f}{\partial z}] t^{1-n}##
But I can't continue with it.

 

[PeroK]

Homework Statement: Supposed f is a function of several variables that satisfies the equation f(tx, ty, tz) =t^{n}f(x, y, z),(t as any number).
Prove:
##xf_{x}+yf_{y}+zf_{z}=nf(x, y, z) ##
Relevant Equations: Partial derivative related formulas

For the first equation:
##f(tx, ty, tz)=f(u, v, w) ##, ##u=tx, v=ty, w=tz##,##k=f(u, v, w) ####
t^{n}f_{x}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}##
As the same calculation
##xf_{x}+yf_{y}+zf_{z}=[\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} +\frac{\partial f}{\partial z}] t^{1-n}##
But I can't continue with it.

Introducing the variables ##u, v, w## looks unnecessary to me. Why not partially differentiate with respect to ##t##?

 

[PeroK]

Prove:
##xf_{x}+yf_{y}+zf_{z}=nf(x, y, z) ##

Note that this equation is somewhat sloppy. The arguments of the function ##f## are given on the right-hand side, but the arguments of the functions ##f_x, f_y## and ##f_z## on the left-hand side are not. More precise and logical would be:$$xf_{x}(x, y, x)+yf_{y}(x, y, z)+zf_{z}(x, y, z)=nf(x, y, z)$$Or, in shorthand:$$xf_{x}+yf_{y}+zf_{z}=nf$$Where the arguments ##(x, y, z)## are understood by default.

 

[fresh_42]

Note that ##n\cdot f(x,y,z) = \left. \dfrac{d}{dt}\right|_{t=1} \left(t^n f(x,y,z)\right)=\left. \dfrac{d}{dt}\right|_{t=1}f(tx,ty,tz).##

 

[KungPeng Zhou]

Note that this equation is somewhat sloppy. The arguments of the function ##f## are given on the right-hand side, but the arguments of the functions ##f_x, f_y## and ##f_z## on the left-hand side are not. More precise and logical would be:$$xf_{x}(x, y, x)+yf_{y}(x, y, z)+zf_{z}(x, y, z)=nf(x, y, z)$$Or, in shorthand:$$xf_{x}+yf_{y}+zf_{z}=nf$$Where the arguments ##(x, y, z)## are understood by default.

Ok, I have solved it. I need to defferential f(tx, ty, tz) with respect to t