$\qquad \left(\dfrac{1}{x^2}\right)' = -\dfrac{2}{x^3}$
Đặt $y = f(x)= \dfrac{1}{x^2}$
$TXD: D = \Bbb R \backslash\{0\}$
Ta có:
$\quad \Delta y = \dfrac{1}{(x + \Delta x)^2} - \dfrac{1}{x^2}= -\dfrac{\Delta x(2x + \Delta x)}{x^2(x + \Delta x)^2}$
Khi đó:
$y' = f'(x)= \lim\limits_{\Delta x \to 0}\dfrac{\Delta y'}{\Delta x}= \lim\limits_{\Delta x \to 0}\left[-\dfrac{\Delta x(2x + \Delta x)}{x^2(x + \Delta x)^2\Delta x}\right] = -\dfrac{2}{x^3}$
Vậy $\left(\dfrac{1}{x^2}\right)' = -\dfrac{2}{x^3}$
