$\frac{\partial V}{\partial x}=\frac{\partial V}{\partial X}\frac{\partial X}{\partial x}+\frac{\partial V}{\partial Y}\frac{\partial Y}{\partial x}$?

Question

I recently started learning partial differentiation and came across a problem in which I don't understand some parts of the given solution.

If $V$ is a homogeneous function in $x$, $y$ of degree $n$, prove that $\frac{\partial V}{\partial x}$,$\frac{\partial V}{\partial y}$ are each a homogeneous function in $x$, $y$ of degree $(n-1)$.

If $V=f(X,Y)$ where $X=\frac{\partial V}{\partial x}$,$Y=\frac{\partial V}{\partial y}$, show that $X\frac{\partial V}{\partial x}+Y\frac{\partial V}{\partial y}=\frac n{n-1}V$

The solution given is this:

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By Euler's theorem, $$x\frac{\partial V}{\partial x}+y\frac{\partial V}{\partial y}=nV\tag1$$

We know that

$$\color{red}{\frac{\partial V}{\partial x}=\frac{\partial V}{\partial X}\frac{\partial X}{\partial x}+\frac{\partial V}{\partial Y}\frac{\partial Y}{\partial x}}$$

$$x\frac{\partial V}{\partial x}=x\frac{\partial V}{\partial X}\frac{\partial X}{\partial x}+x\frac{\partial V}{\partial Y}\frac{\partial Y}{\partial x}\tag2$$

Similarly, $$y\frac{\partial V}{\partial y}=y\frac{\partial V}{\partial X}\frac{\partial X}{\partial y}+y\frac{\partial V}{\partial Y}\frac{\partial Y}{\partial y}\tag3$$

Adding $(2)$ and $(3)$, $$x\frac{\partial V}{\partial x}+y\frac{\partial V}{\partial y}=\left(x\frac{\partial X}{\partial x}+y\frac{\partial X}{\partial y}\right)\frac{\partial V}{\partial X}+\left(x\frac{\partial Y}{\partial x}+y\frac{\partial Y}{\partial y}\right)\frac{\partial V}{\partial Y}\tag4$$

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The rest of the solution is for the second part which I understand but they assume that the first problem is solved.

Which part of the above solution implies that $\frac{\partial V}{\partial x}$,$\frac{\partial V}{\partial y}$ are each a homogeneous function in $x$, $y$ of degree $(n-1)$?

Also I don't understand the part colored in red. I'm not sure if it is too basic but any help is highly appreciated.

Answer 1

The fact that $V$ is a homogeneous function of degree $n$, implies that $\partial_x V, \partial_y V$ are homogeneous functions of degree $n-1$. Indeed, we know that, taken $t>0$, $$V(tx,ty)=t^nV(x,y)$$ Now, deriving w.r.t. $x$, we get $$\frac{\partial V}{\partial x}(tx,ty)\,t=t^n\frac{\partial V}{\partial x}(x,y)\iff\frac{\partial V}{\partial x}(tx,ty)=t^{n-1}\frac{\partial V}{\partial x}$$ For what it concerns the red part, since $V=f(X,Y)$, then using the chain rule, $$\frac{\partial V}{\partial x}=\frac{\partial f}{\partial X}\frac{\partial X}{\partial x}+\frac{\partial f}{\partial Y}\frac{\partial Y}{\partial x}=\frac{\partial V}{\partial X}\frac{\partial X}{\partial x}+\frac{\partial V}{\partial Y}\frac{\partial Y}{\partial x}$$ since $\partial_X V=\partial_X f$.