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