Question

2. Let a and β be multi-indices such that β ≤ a. Prove that
∂^(β) x^(a) = Σ (a,β) β!x^(a - β)
β≤a

Answer 1

Since it is not indicated that $x\in\mathbb{R}^n$ , so I will assume for simplicity that

$$\partial^\beta\left(x^\alpha\right)\equiv d^\beta\left(x^\alpha\right)$$

Remarks: in the condition there is an expression $\left(\alpha,\,\beta\right)$ - this is a binomial coefficient.When solving the problem, I will write it in the traditional form

$$\left(\alpha,\,\beta\right)\equiv\dbinom{\alpha}{\beta}\equiv\frac{\alpha!}{\beta!\cdot(\alpha-\beta)!}$$

( More information: https://en.wikipedia.org/wiki/Binomial_coefficient )

Next, by definition

$$d\left(x^\alpha\right)=\alpha\cdot x^{\alpha-1}\quad\textnormal{Power rule}\\[0.5cm] d^n\left(f\right)=d^{n-1}\left(d\left(f\right)\right)\equiv d^{n-1}\left(f'\right)$$

( More information: https://en.wikipedia.org/wiki/Differentiation_rules )

Then,

$$d^\beta\left(x^\alpha\right)=d^{\beta-1}\left(d\left(x^\alpha\right)\right)= d^{\beta-1}\left(\alpha\cdot x^{\alpha-1}\right)=\\[0.3cm] =\alpha\cdot d^{\beta-2}\left( d\left(x^{\alpha-1}\right)\right)= \alpha(\alpha-1)\cdot d^{\beta-2}\left(x^{\alpha-2}\right)=\\[0.3cm] =\ldots=\alpha(\alpha-1)\cdot\ldots\cdot(\alpha-(\beta-1))\cdot x^{\alpha-\beta}=\\[0.3cm] =\frac{\alpha(\alpha-1)\cdot\ldots\cdot1}{(\alpha-\beta)\cdot\ldots(1)\cdot\beta!}\cdot \beta!\cdot x^{\alpha-\beta}=\frac{\alpha!}{(\alpha-\beta)!\cdot\beta!}\cdot\beta!\cdot x^{\alpha-\beta}$$

Conclusion,

$$\boxed{d^\beta\left(x^\alpha\right)=\frac{\alpha!}{(\alpha-\beta)!\cdot\beta!}\cdot\beta!\cdot x^{\alpha-\beta} \equiv\dbinom{\alpha}{\beta}\cdot\beta!\cdot x^{\alpha-\beta}}$$

Q.E.D.