A function is called homogeneous of degree n if it satisfied the equation f(tx,ty) =t^(n) f(x,y), for all t, where n is a positive integer and f has continuous 2nd order partial derivatives.
If f is homogeneous of degree n, show that df/dx (tx,ty) = t^(n-1) df/dx(x,y)
*Using df/dx for partial derviatives.
The Attempt at a Solution
Basically I've taken the partial derivatives of each side of the definition of homogeneous equation above, applying the chain rule (that's what section the problem is in). What I get is:
d/d(tx) (tx,ty) * t = t^n d/dx(x,y)
which simplifies to df/d(tx) (tx,ty) = t^(n-1) df/dx(x,y).
It feels like I'm done, but I can't figure out-conceptually or otherwise- why the partial derivative with respect to (tx) of f(tx,ty) is the same as the partial derivative with respect to x of f(tx,ty).
Thanks in advance for help!