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Question Number 187155 by TUN last updated on 14/Feb/23

$$f\left(x\right)=x.e^{2x}$$ $$=>f^{\left(n\right)}\left(x\right)=$$ $$$$

Commented by0670322918 last updated on 14/Feb/23

Commented bymr W last updated on 15/Feb/23

$$sir:$$ $$pleasepostyouransweras\mathrm{♮}answer\epsilon ,$$ $$notas\mathrm{♮}comment\epsilon !thanks!$$

Answered by mahdipoor last updated on 14/Feb/23

$$\Rightarrow \Rightarrow f^{\left(0\right)}=a_0{xe}^{2x}+b_0=1{xe}^{2x}+0$$ $$f^{(m-1)}=a_{m-1}{xe}^{2x}+b_{m-1}{e}^{2x}\Rightarrow$$ $$f^{\left(m\right)}=\left(2a_{m-1}\right){xe}^{2x}+(a_{m-1}+2b_{m-1})e^{2x}=a_m{xe}^{2x}+b_m{e}^{2x}$$ $$\Rightarrow \Rightarrow get{f}^{\left(n\right)}=a_n{xe}^{2x}+b_n{e}^{2x}$$ $$a_n=2a_{n-1}=4a_{n-1}=...=2^n{a}_0=2^n$$ $$b_n=a_{n-1}+2b_{n-1}=a_{n-1}+2a_{n-2}+4b_{n-2}=...=$$ $$a_{n-1}+2a_{n-2}+4a_{n-3}+...+2^{n-1}{a}_0+2^n{b}_0=2^{n-1}n$$ $$\Rightarrow \Rightarrow f^{\left(n\right)}=2^n{xe}^{2x}+2^{n-1}{ne}^{2x}$$

Answered by Mathspace last updated on 15/Feb/23

$$f^{\left(n\right)}\left(x\right)=\sum _{k=0}^n{C}_n^k{\left(x\right)}^{\left(k\right)}{\left(e^{2x}\right)}^{n-k}$$ $$=x{\left(e^{2x}\right)}^{\left(n\right)}+C_n^1{\left(e^{2x}\right)}^{(n-1)}$$ $$=x.2^n{e}^{2x}+n{2}^{n-1}{e}^{2x}$$ $$=(x{2}^n+n{2}^{n-1})e^{2x}$$

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