# Partial differentiation of the general function homogeneous of degree n [on hold]

If $$f$$ is homogeneous of degree $$n$$, $$f(tx,ty) = t^{n}f(x,y)$$
show that $$f_{x}(tx,ty) = t^{n-1}f_{x}(x,y)$$

My proof went a little wrong as follow:
$$u=tx, v=ty \quad f_{x}(tx,ty) = \frac{\partial f(u,v)}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f(u,v)}{\partial v}\cdot \frac{\partial v}{\partial x} = f_{u}(u,v) \cdot t$$…….(1)
$$\quad\quad\quad\quad\quad\quad\frac{\partial}{\partial x}(t^nf(x,y))=t^nf_{x}(x,y)$$…….(2)
(1)=(2)$$\qquad f_{u}(u,v)=t^{n-1}f_{x}(x,y)$$
$$\quad \quad \quad\quad \,f_{u}(tx,ty) = t^{n-1}f_{x}(x,y)$$

On the last line, the footnote on the left side is supposed to be $$x$$, however, I get $$u$$.