Question

Dạo hàm y=x.e^{\(\dfrac{-x^2}{2}\)}

Answer 1

\(y'=e^{\dfrac{-x^2}{2}}+x\left(e^{\dfrac{-x^2}{2}}\right)'=e^{\dfrac{-x^2}{2}}+x.e^{\dfrac{-x^2}{2}}.\left(\dfrac{-x^2}{2}\right)'=e^{\dfrac{-x^2}{2}}-x^2.e^{\dfrac{-x^2}{2}}\)

Cách khác: lấy ln 2 vế \(lny=lnx+ln\left(e^{\dfrac{-x^2}{2}}\right)=lnx-\dfrac{x^2}{2}\)

Đạo hàm 2 vế:

\(\dfrac{y'}{y}=\dfrac{1}{x}-x\Rightarrow y'=y\left(\dfrac{1}{x}-x\right)=x.e^{\dfrac{-x^2}{2}}\left(\dfrac{1}{x}-x\right)=e^{\dfrac{-x^2}{2}}-x^2.e^{\dfrac{-x^2}{2}}\)