Cau 6
Ta có $f(x) = \sqrt[3]{x} = x^{\frac{1}{3}}$. Do đó
$f'(x) = \dfrac{1}{3} x^{-\frac{2}{3}} = \dfrac{1}{3} \dfrac{1}{\sqrt[3]{x^2}}$
Do đó
$f'(8) = \dfrac{1}{3}. \dfrac{1}{\sqrt[3]{64}} = \dfrac{1}{3} . \dfrac{1}{4} = \dfrac{1}{12}$
Đáp án B.
Câu 7
Ta có
$f'(x) = \left[ \left( \sqrt{x} - \dfrac{1}{\sqrt{x}} \right)^2 \right]'$
$= 2 \left( \sqrt{x} - \dfrac{1}{\sqrt{x}} \right) . \left( \sqrt{x} - \dfrac{1}{\sqrt{x}} \right)'$
$= 2 \left( \sqrt{x} - \dfrac{1}{\sqrt{x}} \right) . \left( \dfrac{1}{2\sqrt{x}} + \dfrac{1}{2 x\sqrt{x}} \right)$
$= 2 \left( \dfrac{1}{2} + \dfrac{1}{2x} - \dfrac{1}{2x} - \dfrac{1}{2x^2} \right)$
$= 1 - \dfrac{1}{x^2}$
