[Lớp 12] Đạo hàm

X

xuanquynh97

$y=\sqrt[4]{x-2}+\sqrt[4]{4-x}$

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

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

$=\dfrac{1}{4}(x-2)^{\dfrac{-3}{4}}+\dfrac{1}{4}(4-x)^{\dfrac{-3}{4}}$

$=\dfrac{1}{4}.\dfrac{1}{\sqrt[4]{(x-2)^3}}+\dfrac{1}{4}.\dfrac{1}{\sqrt[4]{(4-x)^3}}$
 
You must log in or register to reply here.
Top Bottom