According to textbook we are using at uni, the Euler's theorem which says that if $f(x,y)$ is homogeneous of of degree $k$ then:

$$x f_1'(x,y) + yf_2'(x,y) = kf(x,y)$$

textbook says that this can be proven by differentiating the following with respect to $t$:


This should give us:

$$ x f_1'(tx,ty) + yf_2'(tx,ty) = kt^{k-1}f(x,y)$$

and setting $t=1$ gives us the equation of Euler theorem.

I understand why on right hand side we get $kt^{k-1}f(x,y)$, but I don't understand why chain rule gives us the left hand side expression. If we are differentiating with respect to $t$ why $f_1'$ and $f_2'$ which are derivatives with respect to $x$ and $y$ are there. It is not mentioned that $x$ or $y$ are functions of $t$, so I don't understand why it is not some $f_t'$.

Thanks for any help.

  • $\begingroup$ Seems like a typo, it indeed should be $xf'_{x} (tx,ty) + yf'_{y} (tx,ty)$ before setting t=1. $\endgroup$
    – Tom Ariel
    Nov 14, 2020 at 11:28
  • $\begingroup$ @TomAriel yea, my bad, fixed it $\endgroup$
    – WilliamT
    Nov 14, 2020 at 11:33
  • $\begingroup$ Google the multivariable chain rule. There are many good videos explaining it, and it should clear up the confusion with this question $\endgroup$
    – Ben Martin
    Nov 14, 2020 at 11:56
  • 1
    $\begingroup$ The chain rule says that if f(u, v) is a function of u and v and u and v are themselves functions of t, u(t) and v(t), then $\frac{df}{dt}= \frac{\partial f}{\partial u}\frac{du}{dt}+ \frac{\partial f}{\partial v}\frac{dv}{dt}$. If u= xt and v= yt then that gives $f_t= xf'_1+ yf'_2$. $\endgroup$
    – user247327
    Nov 14, 2020 at 12:58


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