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Question Number 52673 by maxmathsup by imad last updated on 11/Jan/19

$$letf\left(x\right)=(x^n-1)e^{x}determine{f}^{\left(n\right)}\left(x\right)withnintegrnatural$$

Commented by maxmathsup by imad last updated on 13/Jan/19

$$f^{\left(n\right)}\left(x\right)=\sum _{k=0}^n{C}_n^k{(x^n-1)}^{\left(k\right)}{\left(e^x\right)}^{(n-k)}\left(leibnizformulae\right)$$$$=\sum _{k=0}^n{C}_n^k{(x^n-1)}^{\left(k\right)}{e}^x=(x^n-1)e^x+e^x\sum _{k=1}^n{C}_n^k{\left(x^n\right)}^{k}butwehave$$$${\left(x^n\right)}^{\left(1\right)}={nx}^{n-1},{\left(x^n\right)}^{\left(2\right)}=n(n-1)x^{n-2}....{\left(x^n\right)}^{\left(k\right)}=n(n-1)...(n-k+1)x^{n-k}$$$$=\frac{n!}{(n-k)!}{x}^{n-k}\Rightarrow f^{\left(n\right)}\left(x\right)=(x^n-1)e^x+e^x\sum _{k=1}^n\frac{n!}{k!(n-k)!}\frac{n!}{(n-k)!}{x}^{n-k}\Rightarrow$$$$f^{\left(n\right)}\left(x\right)=(x^n-1)e^x+e^x\sum _{k=1}^n\frac{{(n!)}^2}{k!(n-k){!}^2}{x}^{n-k}.$$

Answered by Smail last updated on 11/Jan/19

$$f^{\prime }\left(x\right)=({nx}^{n-1}+x^n-1)e^x$$$$f^{″}\left(x\right)=(n(n-1)x^{n-2}+2{nx}^{n-1}+x^n-1)e^x$$$$f^{\left(3\right)}\left(x\right)=(n(n-1)(n-2)x^{n-3}+3n(n-1)x^{n-2}+3{nx}^{n-1})e^x+f\left(x\right)$$$$f^{\left(4\right)}\left(x\right)=(n\to (n-3)x^{n-4}+4n\to (n-2)x^{n-3}+6n(n-1)x^{n-2}+4{nx}^{n-1})e^x+f\left(x\right)$$$$f^{\left(n\right)}\left(x\right)=(P_{n-0}^n{C}_0^n{x}^0+P_{n-1}^n{C}_1^n{x}^1+P_{n-2}^n{C}_2^n{x}^2+...+P_0^n{C}_n^n{x}^n-1)e^x$$$$f^{\left(n\right)}\left(x\right)=e^x(\underset{k=0}{\sum ^n}{P}_{n-k}^n{C}_k^n{x}^k-1)$$

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