Respuesta :
[tex]\bf \stackrel{\textit{limit de}\textit{finition of a derivative}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ f(x) = 6x + 6x^2\qquad \qquad \lim\limits_{h\to 0}~\cfrac{[6(x+h)+6(x+h)^2]~~-~~[6x+6x^2]}{h} \\\\\\ \lim\limits_{h\to 0}~\cfrac{[6x+6h+6(x^2+2xh+h^2)]~~-~~[6x+6x^2]}{h} \\\\\\ \lim\limits_{h\to 0}~\cfrac{[6x+6h+6x^2+12xh+6h^2]~~-~~[6x+6x^2]}{h}[/tex]
[tex]\bf \lim\limits_{h\to 0}~\cfrac{[~~\begin{matrix} 6x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+6h~~\begin{matrix} +6x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+12xh+6h^2]~~~~\begin{matrix} -6x-6x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{h}[/tex]
[tex]\bf \lim\limits_{h\to 0}~\cfrac{6h+12xh+6h^2}{h}\implies \lim\limits_{h\to 0}~\cfrac{6~~\begin{matrix} h \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ (1+2x+h)}{~~\begin{matrix} h \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ } \\\\\\ \lim\limits_{h\to 0}~6(1+2x+0)\implies 6+12x[/tex]