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Question Number 90630 by jagoll last updated on 25/Apr/20

$$f\left(x\right)={xe}^{-x}$$$$f^{\left(2020\right)}\left(x\right)=$$

Commented by john santu last updated on 25/Apr/20

$$f^{\left(n\right)}\left(x\right)=\{\begin{array}{l}(n-x).e^{-x},nodd\\ (x-n).e^{-x},neven\end{array}$$$$f^{\left(2020\right)}\left(x\right)=(x-2020).e^{-x}$$

Answered by MWSuSon last updated on 25/Apr/20

$$I^{\prime }massumingyoumean2020thderivative$$$$usingLeibnitztheorem{f}^{\left(n\right)}\left(x\right)=[{(-1)}^n{xe}^{-x}+n{(-1)}^{n-1}{e}^{-x}]$$$$f^{\left(2020\right)}=[{(-1)}^{2020}{xe}^{-x}+2020{(-1)}^{2020-1}{e}^{-x}]$$$$f^{\left(2020\right)}=[{xe}^{-x}-2020e^{-x}]=e^{-x}[x-2020]$$

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