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Câu hỏi: Tính \(\varphi '\left( 2 \right),\) biết rằng \(\varphi \left( x \right) = {{\left({x - 2} \right)\left({8 - x} \right)} \over {{x^2}}}.\)
Phương pháp giải
Tính \(\varphi '\left( x \right)\) rồi thay x=2 vào.
Lời giải chi tiết
\(\begin{array}{l}
\varphi \left(x \right) = \dfrac{{\left({x - 2} \right)\left({8 - x} \right)}}{{{x^2}}}\\
= \dfrac{{ - {x^2} + 10x - 16}}{{{x^2}}}\\
= - 1 + \dfrac{{10}}{x} - \dfrac{{16}}{{{x^2}}}\\
\Rightarrow \varphi '\left(x \right) = - \dfrac{{10}}{{{x^2}}} - \dfrac{{ - 16.\left({{x^2}} \right)'}}{{{x^4}}}\\
= - \dfrac{{10}}{{{x^2}}} + \dfrac{{16.2x}}{{{x^4}}} = - \dfrac{{10}}{{{x^2}}} + \dfrac{{32}}{{{x^3}}}\\
\Rightarrow \varphi '\left(2 \right) = - \dfrac{{10}}{{{2^2}}} + \dfrac{{32}}{{{2^3}}} = \dfrac{3}{2}
\end{array}\)
Tính \(\varphi '\left( x \right)\) rồi thay x=2 vào.
Lời giải chi tiết
\(\begin{array}{l}
\varphi \left(x \right) = \dfrac{{\left({x - 2} \right)\left({8 - x} \right)}}{{{x^2}}}\\
= \dfrac{{ - {x^2} + 10x - 16}}{{{x^2}}}\\
= - 1 + \dfrac{{10}}{x} - \dfrac{{16}}{{{x^2}}}\\
\Rightarrow \varphi '\left(x \right) = - \dfrac{{10}}{{{x^2}}} - \dfrac{{ - 16.\left({{x^2}} \right)'}}{{{x^4}}}\\
= - \dfrac{{10}}{{{x^2}}} + \dfrac{{16.2x}}{{{x^4}}} = - \dfrac{{10}}{{{x^2}}} + \dfrac{{32}}{{{x^3}}}\\
\Rightarrow \varphi '\left(2 \right) = - \dfrac{{10}}{{{2^2}}} + \dfrac{{32}}{{{2^3}}} = \dfrac{3}{2}
\end{array}\)