Bài tập 10 trang 195 SGK Toán 11 NC

a) Với x₀ ∈ R ta có:

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

Suy ra f′(3) = 27, f′(−4) = 48

b) Với x₀ > 0, ta có:

\(\begin{array}{l}
f\prime ({x_0}) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f(x) - f({x_0})}}{{x - {x_0}}}\\
 = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sqrt x  - \sqrt {{x_0}} }}{{x - {x_0}}}\\
 = \mathop {\lim }\limits_{x \to {x_0}} \frac{1}{{\sqrt x  + \sqrt {{x_0}} }} = \frac{1}{{2\sqrt {{x_0}} }}\\
 \Rightarrow f\prime (1) = \frac{1}{2},f\prime (9) = \frac{1}{6}
\end{array}\)

-- Mod Toán 11 HỌC247