# Confusion about derivation of Euler's theorem

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 $$t$$:

$$f(tx,ty)=t^kf(x,y)$$

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.

• Seems like a typo, it indeed should be $xf'_{x} (tx,ty) + yf'_{y} (tx,ty)$ before setting t=1. Nov 14, 2020 at 11:28
• @TomAriel yea, my bad, fixed it Nov 14, 2020 at 11:33
• Google the multivariable chain rule. There are many good videos explaining it, and it should clear up the confusion with this question Nov 14, 2020 at 11:56
• 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 $f_t= xf'_1+ yf'_2$. Nov 14, 2020 at 12:58