1.1 Statement

If $f$ and $g$ are differentiable at $x$, and $\alpha,\beta$ are constants, then

$$ \frac{d}{dx}\bigl[\alpha\, f(x) + \beta\, g(x)\bigr] = \alpha\, f'(x) + \beta\, g'(x). $$

In operator notation: $D(\alpha f + \beta g) = \alpha\,Df + \beta\,Dg$. This says the derivative operator is linear.

1.2 Proof from the Definition

Constant-multiple rule. Let $c$ be a constant. Then:

$$ \frac{d}{dx}[c\,f(x)] = \lim_{h\to 0}\frac{c\,f(x+h)-c\,f(x)}{h} = c\,\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = c\,f'(x). $$

We simply factor the constant out of the limit — justified by the constant-multiple law for limits.

Sum rule.

$$ \frac{d}{dx}[f(x)+g(x)] = \lim_{h\to 0}\frac{[f(x+h)+g(x+h)]-[f(x)+g(x)]}{h} $$ $$ = \lim_{h\to 0}\left[\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}\right] = f'(x)+g'(x). $$

Here we used the sum law for limits.

Combining these two results immediately gives the full linearity property.

1.3 Extension to Finite Linear Combinations

By induction, for any $n$ differentiable functions $f_1,\dots,f_n$ and constants $c_1,\dots,c_n$:

$$ \frac{d}{dx}\left[\sum_{k=1}^{n} c_k\,f_k(x)\right] = \sum_{k=1}^{n} c_k\,f_k'(x). $$

1.4 Immediate Consequence — Differentiating Polynomials Term by Term

Every polynomial $p(x)=a_n x^n + a_{n-1}x^{n-1}+\cdots+a_1 x + a_0$ is a finite linear combination of power functions. Using linearity together with the power rule $\frac{d}{dx}[x^k]=k\,x^{k-1}$:

$$ p'(x) = n\,a_n\,x^{n-1} + (n-1)\,a_{n-1}\,x^{n-2} + \cdots + a_1. $$