Cho em hỏi đao hàm của y =x căn bâc hai của 4-x tính sao vậy ak 05/09/2021 Bởi Maya Cho em hỏi đao hàm của y =x căn bâc hai của 4-x tính sao vậy ak
$y=x\sqrt{4-x}$ $y’=x’\sqrt{4-x}+x.(\sqrt{4-x})’$ $=\sqrt{4-x}+x.\dfrac{(4-x)’}{2\sqrt{4-x}}$ $=\sqrt{4-x}-x.\dfrac{1}{2\sqrt{4-x}}$ $=\dfrac{4-x-x}{2\sqrt{4-x}}$ $=\dfrac{2-x}{\sqrt{4-x}}$ Bình luận
Ta xét $y = x\sqrt{4-x}$ Khi đó $y’ = (x.\sqrt{4-x})’$ $= [x.(4-x)^{\frac{1}{2}}]’$ $= x’ . (4-x)^{\frac{1}{2}} + x . [(4-x)^{\frac{1}{2}}]’$ $= 1.(4-x)^{\frac{1}{2}} + x . \dfrac{1}{2} . (4-x)^{-\frac{1}{2}} . (4-x)’$ $= \sqrt{4-x} + x . \dfrac{-1}{2(4-x)^{\frac{1}{2}}}$ $= \sqrt{4-x} -\dfrac{x}{2\sqrt{4-x}}$ Vậy $(x\sqrt{4-x})’ = \sqrt{4-x} -\dfrac{x}{2\sqrt{4-x}}$. Bình luận
$y=x\sqrt{4-x}$
$y’=x’\sqrt{4-x}+x.(\sqrt{4-x})’$
$=\sqrt{4-x}+x.\dfrac{(4-x)’}{2\sqrt{4-x}}$
$=\sqrt{4-x}-x.\dfrac{1}{2\sqrt{4-x}}$
$=\dfrac{4-x-x}{2\sqrt{4-x}}$
$=\dfrac{2-x}{\sqrt{4-x}}$
Ta xét
$y = x\sqrt{4-x}$
Khi đó
$y’ = (x.\sqrt{4-x})’$
$= [x.(4-x)^{\frac{1}{2}}]’$
$= x’ . (4-x)^{\frac{1}{2}} + x . [(4-x)^{\frac{1}{2}}]’$
$= 1.(4-x)^{\frac{1}{2}} + x . \dfrac{1}{2} . (4-x)^{-\frac{1}{2}} . (4-x)’$
$= \sqrt{4-x} + x . \dfrac{-1}{2(4-x)^{\frac{1}{2}}}$
$= \sqrt{4-x} -\dfrac{x}{2\sqrt{4-x}}$
Vậy $(x\sqrt{4-x})’ = \sqrt{4-x} -\dfrac{x}{2\sqrt{4-x}}$.