How to show the following operator is linear?

Question

Let us consider the linear vector space $P_n(I)$ of all polynomials defined over an interval $I$ is subset Real Number Set with degrees less than the positive integer $n$ . A basis for this space is given by $\{1, x, x^2, ... , x^{n-1}\}$. We denote the differentiation with respect to the variable $x$ by the operator $D$. How can we show that $D : P_n(I)\rightarrow P_{n-1}(I)$ is a linear operator?

I have no ideas. Could you give some hints?

Linearity:

where $T:X \rightarrow Y$ is an operator, $u,v\in X; α\in K$

$T(u+v) = T(u) + T(v)$

$T(αv) = αT(v)$

Answer 1

Hint 1: Think how you would go about differentiating $f(x)=x^2+x^3$ or $g(x)=a\cdot x^9, \ \ a\in \Bbb{R}$.

Hint 2. The derivative is really a limit. In that way, the reasoning relies upon two limit properties.

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Assume all limits exist. Here $u_i=u_i(x)$.

Theorem 1. $$\lim\left(\sum_{i=1}^nu_i\right)=\sum_{i=1}^n\lim u_i$$

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Theorem 2. $$\lim[a\cdot u (x)]=a\cdot\lim u(x), \ \ a\in\Bbb{R}$$