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Question Number 133194 by Algoritm last updated on 19/Feb/21

Commented by Algoritm last updated on 19/Feb/21

$$\frac{{\mathrm{d}}^{\mathrm{n}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{n}}}=?$$

Commented by mr W last updated on 20/Feb/21

$$verystrange!$$$$doesthisguyreallyhavemany$$$$suportersorhejustusesmany$$$$differentIDs?eachtimewhenhe$$$$haspostedaquestion,nomatterhow$$$$goodorbadthequestionis,thepost$$$$is\mathit{i}\mathit{m}\mathit{m}\mathit{e}\mathit{d}\mathit{i}\mathit{a}\mathit{t}\mathit{e}\mathit{l}\mathit{y}likedoneormore$$$$times.i{can}^{\prime }tbeleavethattheselikes$$$$areseriouslymeantand{don}^{\prime }t$$$$understandwhattheyaregoodfor.$$

Answered by Olaf last updated on 19/Feb/21

$$f\left(x\right)=\ln (x^2+1)=\ln (x-i)+\ln (x+i)$$$$f^{\prime }\left(x\right)=\frac{2x}{1+x^2}=\frac{1}{x-i}+\frac{1}{x+i}$$$$f^{\left(n\right)}=\frac{{(-1)}^{n-1}(n-1)!}{{(x-i)}^n}+\frac{{(-1)}^{n-1}(n-1)!}{{(x+i)}^n}$$$$f^{\left(n\right)}={(-1)}^{n-1}(n-1)![\frac{1}{{(x-i)}^n}+\frac{1}{{(x+i)}^n}]$$$$f^{\left(n\right)}={(-1)}^n(n-1)!\left[\frac{{(x+i)}^n+{(x-i)}^n}{{(x^2+1)}^n}\right]$$$$f^{\left(n\right)}={(-1)}^{n-1}(n-1)!\left[\frac{\underset{k=0}{\sum ^n}{\mathrm{C}}_k^n{x}^k{i}^{n-k}+\underset{k=0}{\sum ^n}{\mathrm{C}}_k^n{x}^k{(-i)}^{n-k}}{{(x^2+1)}^n}\right]$$$$f^{\left(n\right)}={(-1)}^{n-1}(n-1)!\left[\frac{\underset{k=0}{\sum ^n}{\mathrm{C}}_k^n{x}^{n-k}(i^k+{(-i)}^k)}{{(x^2+1)}^n}\right]$$$$f^{\left(n\right)}={(-1)}^{n-1}(n-1)!\left[\frac{\underset{p=0}{\sum ^{\lfloor \frac{n}{2}\rfloor }}{\mathrm{C}}_{2p}^n{x}^{n-2p}{i}^{2p}(1+{(-1)}^{2p})}{{(x^2+1)}^n}\right]$$$$f^{\left(n\right)}={(-1)}^{n-1}(n-1)!\left[\frac{\underset{p=0}{\sum ^{\lfloor \frac{n}{2}\rfloor }}{\mathrm{C}}_{2p}^n{x}^{n-2p}{(-1)}^p\times 2}{{(x^2+1)}^n}\right]$$$$f^{\left(n\right)}=2{(-1)}^{n-1}(n-1)!\left[\frac{\underset{p=0}{\sum ^{\lfloor \frac{n}{2}\rfloor }}{(-1)}^p{\mathrm{C}}_{2p}^n{x}^{n-2p}}{{(x^2+1)}^n}\right]$$

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