Let n be a positive integer. A function F is called honogeneous of degree n if it satisfies the equation F(tx,ty) = tnF(x,y) for all real t. Suppose f(x,y) has continuous second-order partial derivatives.
Show that if F is homogeneous of degree n, then
x*F_x + y*F_y = n*F(x,y), where F_x,F_y are the partial derivatives
with respect to x,y.
The Attempt at a Solution
Suppose I let u=tx and v=ty. Then,
F_t = F_u * x + F_v * y,
which should then be equal to
n*t^(n-1) * F(x,y).
This, I think, almost looks like what I want to prove. Dividing by
t^(n-1) gives n*F(x,y) and (F_u * x + F_v * y)/(t^(n-1)), which I want
to rewrite as
x*F_x + y*F_y.
But I have no idea if/why this should be true. Am I thinking about this correctly, or have I done it the wrong way?