How to prove that $∇f(x)\cdot x = tf(x) $? Suppose we have $ t ∈ \mathbb R $ and $ f : \mathbb R^n → \mathbb R $ which satisfies $ f(λx) = λ^tf(x) $ for $ x ∈ \mathbb R^n $, $ λ ∈ \mathbb R $. 
I have two questions;
How would I prove that $ ∇f(x)\cdot x = tf(x) $ ?
And how can I verify this formula in case $ n = 2 $ for the
function given by $ f(x, y) = x^2 + 2xy + y^2 $ ?
My initial idea was to use the fact that $f(x) = \frac{f(λx)}{(λ^t)} $ and then find $ ∇f(x) $ of this.
I'm struggling to find a solution that I can follow step by step, any help would be appreciated.
 A: First, $\lambda$ should be positive. Second, to derive Euler's formula (this is how it's called), just take the derivative with respect to $\lambda$ in both sides of $f(\lambda x) = \lambda ^t f(x)$ and use the chain rule:


*

*the left-hand side gives $\sum _{i = 1} ^n \dfrac {\partial f} {\partial x_i} (\lambda x) \ x_i = (\nabla f)(\lambda x) \cdot x$

*the right-hand side gives $t \lambda ^{t-1} f(x)$.
Being both sides of an equality, these expressions must be equal, so evaluate them both in $\lambda =1$. This is it.
In the concrete case of $n=2$ and $f(x,y) = x^2 + 2xy + y^2$, notice that
$$f( \lambda x, \lambda y) = (\lambda x)^2 + 2( \lambda x) (\lambda y) + (\lambda y)^2 = \lambda^2 (x^2 + 2xy + y^2) = \lambda ^2 f(x,y) ,$$
so in this case $t=2$. You get $\dfrac {\partial f} {\partial x} (x,y) x + \dfrac {\partial f} {\partial y} (x,y) y = 2 f(x,y)$.
A: Hint: differentiate your equation $f(λx) = λ^tf(x)$ w.r.t. $\lambda$ and then set $\lambda=1$.
A: Let $x\in \Bbb R^n$ be fixed and define $$g_x(\lambda)=f(\lambda x)-\lambda^tf(x)$$
Then, $g(\lambda)=0$ for every $\lambda$, therefore $g$ is constant. Thus $g'(\lambda)=0$ for every $\lambda$. In particular,
$$0=g'(1)=\langle x,\nabla f(x)\rangle-tf(x).$$

For your second question: 
Note that $\nabla f(x,y) = (2x+y,2y+x)^T$ and thus $$\langle (x,y),\nabla f(x,y)\rangle=x(2x+y)+y(2y+x)=2f(x,y)$$
