Calculate $\partial^\alpha(x_j f(x_1,...,x_n))$ Just a very basic computational question:

Let $\alpha \in \mathbb{N}^n_0$. Calculate $\partial^\alpha(x_j f(x_1,...,x_n))$ where $f :\mathbb{R}^n \to \mathbb{C}$ is smooth and $j \in \{1,...,n\}$.

Using Leibniz I get,
$\partial^\alpha(x_j f(x_1,...,x_n))= x_j\partial^\alpha f+ \alpha_j \partial^{\alpha-e_j}f$.
But my book says
$\partial^\alpha(x_j f(x_1,...,x_n))= x_j\partial^\alpha f+ \alpha_j \partial^{\alpha-\alpha_je_j}f$.
The difference being in the exponents of the partial derivatives in the second terms.
Where did I go wrong?
 A: Take $\alpha_j=2$. Then freezing all the other variables and calling $t$ the $x_j$ variable, you have $$\frac{d^2}{dt^2}(tf(t))=\frac{d}{dt}(1f(t)+tf’(t))=f’(t)+1f’(t)+f’’(t).$$ Following the formula in your book you should get $f$ instead of $f’$. There is a mistake in your book.
There is a formula for the product rule of two functions using multi-indexes. product
Now, $\partial^\beta(x_j)\ne 0$ only if $\beta=0$ or $\beta=e_j$ and so you find exactly your formula.
A: Another example where the book's formula fails: Let $n=2$ and take $f(x) =  x_1x_2^2.$ Then $x_1f(x) = x_1^2x_2^2.$ Clearly $D^{(2,2)}(x_1f) = 4.$ But your book says the answer is
$$x_1D^{(2,2)}f +2D^{(2,2)-(2,0)}(x_1x_2^2)= 0 +2D^{(0,2)}(x_1x_2^2) = 2\cdot 2x_1 = 4x_1,$$
which is incorrect. Your formula gives the correct answer here. I think your formula is valid in all cases, as long as we agree $\alpha_j \partial^{\alpha-e_j}f =0$ in the case $\alpha_j=0.$ (There's a question of what $D^{\alpha-e_j}f $ means if $\alpha_j=0.$)
Let's check the formula using Leibnitz. If $\alpha_j=0,$ then $D^\alpha x_jf= x_jD^\alpha f,$ which is what your formula gives. If $\alpha_j>0,$ then
$$D^\alpha (x_jf) = D^{\alpha-\alpha_je_j}[D^{\alpha_je_j} (x_jf)].$$
By Leibnitz, the expression inside the brackets equals
$$\sum_{k=0}^{\alpha_j}\binom{\alpha_j}{k}(D^{ke_j}x_j)(D^{(\alpha_j -k)e_j}f) = x_jD^{\alpha_je_j}f + \alpha_jD^{(\alpha_j -1)e_j}f.$$
The first term on the right is the $k=0$ term; the second term is the $k=1$ term. For $k>1,$ $D^{ke_j}x_j=0.$
Applying $D^{\alpha-\alpha_je_j}$ to this gives
$$x_jD^{\alpha}f  +\alpha_jD^{\alpha -e_j}f,$$
and there's your formula.
