Help to understand the proof of partial derivatives of homogeneous functions I found this short proof that says the partial derivaties of homogenous functions of degree $k$ is homogeneous of degree $k-1$. Here is the proof in its entirety:

I am lost at the very first step of the proof which says to differentiate with respect to $x_i$ both sides of the equation:
$f(tx_1,tx_2,\dots,tx_n)=t^kf(x_1,x_2,\dots,x_n)$
Based on my understanding of partial derivates, if I were to differentiate the left hand side, I will get this:
$tf'_i(x_1,x_2,\dots,x_n)$
Which is not the same as what the proof says it should be.
Please advise.
 A: To be sure, do each step carefully. Write 
$$ \phi_t(x)=(tx_1\ldots tx_n).$$
Therefore you are trying to compute 
$$\frac{\partial}{\partial x_i} \left( f\circ \phi_t\right)(x_1\ldots x_n).$$
Now use the chain rule. 
A: Another way to see it (from here):

Theorem 2: If $f : R^n_{++} \to R$ is continuously differentiable and homogeneous of degree $\alpha$, then each partial derivative $f_i$ is homogeneous of degree $\alpha − 1$.
Proof. For fixed $x ∈ R^n_{++}$ and $\lambda > 0$, define each $g_i, h_i : (−x_i,\infty) \to R$ by $g_i(t) = f(\lambda (x + e_it))$ and $h_i(t) = \lambda^{\alpha}f(x + e_it)$ Then the homogeneity of f implies
$$g_i(t) = f(\lambda (x + te_i)) = \lambda^\alpha f(x + te_i) = h_i(t)$$
and therefore
$$g'_i(t) = h'_i(t)\quad \mbox{for all}\quad t \in (−x_i,\infty)$$
But
$$g'_i(0) = D f (\lambda x) \cdot \lambda e_i = \lambda f_i(\lambda x)$$
$$h'_i(0) = \lambda^\alpha Df(x) \cdot e_i = \lambda^\alpha f_i(x)$$
So
$$f_i (\lambda x) = \lambda^{\alpha−1}f_i(x).$$
$\blacksquare$

And to answer your question directly i expand the formula you have doubts in:
$$t\frac{\delta f(tx_1,\dots,tx_n)}{\delta (tx_i)} = t^k \frac{\delta f(x_1,\dots,x_n)}{\delta x_i}$$
