Question

VALE 40 PONTOS......alguém sabe o resultado da derivada de y = x^2(x ao quadrado)-x+2. --- PELO LIMITE.

Answer 1

$\underset{h\to0}\lim\ \frac{((x+h)^2-(x+h)+2)-(x^2-x+2)}{h}\\\\\underset{h\to0}\lim\ \frac{x^2+2xh+h^2-x-h+2-x^2+x-2}{h}\\\\\underset{h\to0}\lim\ \frac{2xh+h^2-h}{h}\\\\\underset{h\to0}\lim\ 2x+h-1\\\\2x+0-1\\\\\boxed{2x-1}$

Answer 2

$\frac{d}{dx}(x^2-x+2) = f'(x)$
$f'(x) = \lim_{\Delta x \to 0 }\frac{f(x+\Delta x) - f(x)}{\Delta x}$
$f'(x) = \lim_{\Delta x \to 0 }\frac{(x+\Delta x)^2-(x+\Delta x) + 2 -(x^2-x+2)}{\Delta x}$
$f'(x) = \lim_{\Delta x \to 0 }\frac{x^2 + 2x \Delta x + \Delta x^2 -x - \Delta x +2 -x^2+x-2}{\Delta x}$
$f'(x) = \lim_{\Delta x \to 0 }\frac{2x \Delta x + \Delta x^2 - \Delta x}{\Delta x}$
$f'(x) = \lim_{\Delta x \to 0 } 2x + \Delta x -1 = 2x-1$

$f'(x) = 2x - 1$