[Lớp 12] Đạo hàm

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xuanquynh97

y=x−24+4−x4y=\sqrt[4]{x-2}+\sqrt[4]{4-x}

=(x−2)14+(4−x)14=(x-2)^{\dfrac{1}{4}}+(4-x)^{\dfrac{1}{4}}

\Rightarrow y′=14(x−2)14−1+14(4−x)14−1y'=\dfrac{1}{4}(x-2)^{\dfrac{1}{4}-1}+\dfrac{1}{4}(4-x)^{\dfrac{1}{4}-1}

=14(x−2)−34+14(4−x)−34=\dfrac{1}{4}(x-2)^{\dfrac{-3}{4}}+\dfrac{1}{4}(4-x)^{\dfrac{-3}{4}}

=14.1(x−2)34+14.1(4−x)34=\dfrac{1}{4}.\dfrac{1}{\sqrt[4]{(x-2)^3}}+\dfrac{1}{4}.\dfrac{1}{\sqrt[4]{(4-x)^3}}
 
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