# $f(x,y)$ is homogeneous function of degree 3, and $f'_x(2,1)=1$, $f(1,0.5)=2$, Compute: $f'_y(4,2)$

I'd love your help with the following question:

Let $f(x,y)$ be a homogeneous function of degree 3. I'm trying to compute $f'_y(4,2)$, assuming that $f'_x(2,1)=1$, $f(1,0.5)=2$. by knowing that $f$ is a 3 degree homogeneous, I also know that $f(4,2)=128$, $f(2,1)=16$. I tried to use the definition of the partial derivatives, but it didn't work for me.

Any suggestion?

Thanks a lot!

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Does it also mean that $f'_x$ is homogeneous? – Jozef Jan 23 '12 at 18:58
A more usual notation for partial derivative with respect to $x$ is $f_x$. No ${}'$ is needed. – Robert Israel Jan 23 '12 at 19:27
Consider the function $g(t)=f(2t,t)$. Due to homeogenity, it must have the form $g(t)=at^3$, so you have information enough to compute $a$. Therefore you can compute $g'(1)$ which by the chain rule equals $2f_x(2,1)+f_y(2,1)$. Since you know $f_x(2,1)$ you can compute $f_y(2,1)$ and by appropriate scaling of this you'll find $f_y(4,2)$. – Henning Makholm Jan 23 '12 at 19:39

Note that $$f_x(x,y) = \frac{\partial}{\partial x} (x^3 f(1,y/x)) = 3 x^2 f(1,y/x) - x y f_y(1,y/x) = (3/x) f(x,y) - (y/x) f_y(x,y)$$
Let $f$ be a homogeneous function of degree $n$. By Euler's Theorem you know that $xf_x+yf_y=nf$. Also, you have that the partial derivatives of $f$ are homogeneous functions of degree $n-1$. Then $4f_x(4,2)+2f_y(4,2)=3f(4,2)$ and finally $f_y(4,2)=184$.