Tính g′(1), biết rằng \(g(x) = \frac{1}{x} + \frac{1}{{\sqrt x }} + {x^2}\)
Ta có:
\(g'(x) = - \frac{1}{{{x^2}}} - \frac{1}{{2x\sqrt x }} + 2x\)
Suy ra \(g'(1) = - 1 - \frac{1}{2} + 2 = \frac{1}{2}\)
-- Mod Toán 11
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