Let $f : \mathbb R^2 →\mathbb R$ be a differentiable function. Answer the following questions:
- Assume that $∀(x, y) ∈$ $\mathbb R^2$ and $∀t ∈$ $\mathbb R$, $f(x + t, y + t) = f(x, y)$.
Show that $\frac{∂f(x, y)}{∂x} + \frac{∂f(x, y)}{∂y} = 0$.
- Assume that $f$ is homogeneous with degree $n$, i.e $∀(x, y) ∈$ $\mathbb R^2$ and $∀t ∈$ $\mathbb R$, $f(tx, ty) = t^n f(x, y)$.
Show that $x\frac{∂f(x, y)}{∂x} + y\frac{∂f(x, y)}{∂y} = nf(x, y)$
I tried working on these two problems for more than an hour but I still didn't get the correct answer. In part 1,I let $u=x+t$ and $v=y+t$, and so by the chain rule, I got that $\frac{∂f(x, y)}{∂x} =\frac{∂f(x, y)}{∂u}. \frac{∂u}{∂x} = \frac{∂f(x, y)}{∂u}$
And $\frac{∂f(x, y)}{∂y} = \frac{∂f(x, y)}{∂v}. \frac{∂v}{∂y} = \frac{∂f(x, y)}{∂v}$. But how to continue from there?
In part 2, I let $u=tx$ and $v=ty$ and so by chain rule,$\frac{∂f(x, y)}{∂x} =\frac{∂f(x, y)}{∂u}. \frac{∂u}{∂x} + \frac{∂f(x, y)}{∂v}. \frac{∂v}{∂x} = t. \frac{∂f}{∂u}$
And $\frac{∂f(x, y)}{∂y} =\frac{∂f(x, y)}{∂u}. \frac{∂u}{∂y} + \frac{∂f(x, y)}{∂v}. \frac{∂v}{∂y} = t. \frac{∂f}{∂v}$. But how to continue from there?
Any help in the 2 parts please?