Bài 1 trang 63 SGK Toán 11 Tập 2 Cánh diều - CD

Ta có:

\(\begin{array}{l}\Delta x = x - {x_0} = x - 1\\\Delta y = f({x_0} + \Delta x) - f({x_0}) = f(x) - f(1)\\\mathop {\lim }\limits_{x \to 1} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{x \to 1} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^3} - 1 - (3 - 1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^3} - 3}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{3(x - 1)({x^2} + x + 1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} (3({x^2} + x + 1)) = 9\end{array}\)

Vậy \(f'(1) = 9\)

-- Mod Toán 11 HỌC247