Partial derivatives in the proof for $\frac {\partial}{\partial x} f(tx,ty)= t^{n-1} \frac{\partial }{\partial x}f(tx,ty)$

Question

For a homogenous function $f$ of degree n it holds that $$\frac {\partial}{\partial x} f(tx,ty)= t^{n-1} \frac {\partial}{\partial x}f(tx,ty)$$ which can be seen with the chain rule:

$$ \frac {\partial}{\partial (x)} f(tx,ty) = \frac {\partial}{\partial (x)} t^n f(x,y) $$ $$\Leftrightarrow \frac {\partial}{\partial (tx)} f(tx,ty) \cdot \frac {\partial (tx)}{\partial x} + \frac {\partial}{\partial (ty)} f(tx,ty) \cdot \frac {\partial (ty)}{\partial x} = t^n \frac {\partial}{\partial (x)} f(x,y) $$ $$\Leftrightarrow\frac {\partial}{\partial (x)} f(tx,ty) \cdot t= t^n \frac {\partial}{\partial (x)} f(x,y) $$

When you differentiate with respect to $x$, do you just write $f(tx,ty)$ as a function of $x$ holding $t$ and $y$ fixed? Because I've only ever encountered it in the form $f \circ (x,y)$ where $x,y$ are functions of $t$.

My second question, why does $\frac {\partial}{\partial (tx)} f(tx,ty)$ suddenly become $\frac {\partial}{\partial (x)} f(tx,ty)$ in the last line?

Thank you for your help!

Answer 1

In the last line it must be $\frac {\partial}{\partial (tx)} f(tx,ty)$ otherwise it contradicts the original equation in the first line (i.e. extra $t$ is multiplied on the LHS).

An example: Consider: $f(x,y)=x^{n/3}y^{2n/3}$. Then: $$f(5x,5y)=5^nf(x,y) \Rightarrow \\ (5x)^{n/3}\cdot (5y)^{2n/3}=5^n\cdot x^{n/3}y^{2n/3} \Rightarrow \\ \frac{\partial}{\partial x}\left((5x)^{n/3}\cdot (5y)^{2n/3}\right)=\frac{\partial}{\partial x}\left(5^n\cdot x^{n/3}y^{2n/3}\right) \Rightarrow \\ \frac{\partial}{\partial (5x)}\left((5x)^{n/3}\cdot (5y)^{2n/3}\right)\cdot\underbrace{\frac{\partial}{\partial x}(5x)}_{=5} +\frac{\partial}{\partial (5y)}\left((5x)^{n/3}\cdot (5y)^{2n/3}\right)\cdot \underbrace{\frac{\partial}{\partial x}(5y)}_{=0}=\\ 5^n\cdot\frac{\partial}{\partial x}\left(x^{n/3}y^{2n/3}\right) \Rightarrow \\ \frac{\partial}{\partial (5x)}\left((5x)^{n/3}\cdot (5y)^{2n/3}\right)\cdot 5=5^n\cdot\frac{\partial}{\partial x}\left(x^{n/3}y^{2n/3}\right) \Rightarrow \\$$

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$$\frac{\partial}{\partial (5x)}\left((5x)^{n/3}\right)\cdot (5y)^{2n/3}\cdot 5=5^n\cdot y^{2n/3}\cdot \frac{\partial}{\partial x}\left(x^{n/3}\right) \Rightarrow \\ \frac{n}{3}(5x)^{n/3-1}\cdot (5y)^{2n/3}\cdot 5=5^n\cdot y^{2n/3}\cdot \frac{n}{3}x^{n/3-1} \Rightarrow \\ 5^{n/3-1+2n/3+1}=5^n.$$

Answer 2

Yes, when you differentiate with respect to $x$ you are keeping $t$ and $y$ fixed and so you are taking the derivative of the function $g(x)=f(tx,ty)$. You have $g'(x)=t\frac{\partial f}{\partial x}(tx,ty)$ (if you had to differentiate $f(5x)$ you would get $5\frac{\partial f}{\partial x}(5x)$). On the other hand, since $f(tx,ty)=t^nf(x,y)$ if you differentiate the function $h(x)=t^nf(x,y)$ (so again $t$ and $y$ are fixed) with respect to $x$ you get $h'(x)=t^n\frac{\partial f}{\partial x}(x,y)$. Since $g(x)=h(x)$ you have that $g'(x)=h'(x)$ and so by equating the two you get $t\frac{\partial f}{\partial x}(tx,ty)=g'(x)=h'(x)=t^n\frac{\partial f}{\partial x}(x,y)$, that is, $\frac{\partial f}{\partial x}(tx,ty)=t^{n-1}\frac{\partial f}{\partial x}(x,y)$.