Tính \(\varphi '(2)\), biết rằng \(\varphi (x) = \frac{{(x - 2)(8 - x)}}{{{x^2}}}\)
Ta có:
\(\varphi \left( x \right) = \frac{{\left( {x - 2} \right)\left( {8 - x} \right)}}{{{x^2}}} = \frac{{ - {x^2} + 10x - 16}}{{{x^2}}} = - 1 + \frac{{10}}{x} - \frac{{16}}{{{x^2}}} \Rightarrow \varphi '\left( x \right) = - \frac{{10}}{{{x^2}}} + \frac{{32}}{{{x^3}}}\)
Vậy \(\varphi '(2) = - \frac{{10}}{4} + \frac{{32}}{8} = \frac{3}{2}\)
-- Mod Toán 11
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