a. Với \(x_0\in\mathbb R\) ta có:
\(\eqalign{ & f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} {{f\left( x \right) - f\left( {{x_0}} \right)} \over {x - {x_0}}} \cr & = \mathop {\lim }\limits_{x \to {x_0}} {{{x^3} - x_0^3} \over {x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \left( {x+ x{x_0} + x_0^2} \right) = 3x_0^2 \cr} \)
Suy ra \(f'\left( 3 \right) = 27,f'\left( { - 4} \right) = 48\)
b. Với \(x_0> 0\) ta có :
\(\eqalign{ & f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} {{f\left( x \right) - f\left( {{x_0}} \right)} \over {x - {x_0}}} \cr & = \mathop {\lim }\limits_{x \to {x_0}} {{\sqrt x - \sqrt {{x_0}} } \over {x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} {1 \over {\sqrt x + \sqrt {{x_0}} }} = {1 \over {2\sqrt {{x_0}} }} \cr} \)
Suy ra: \(f'\left( 1 \right) = {1 \over 2},f'\left( 9 \right) = {1 \over 6}\)
