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Question Number 93732 by Ar Brandon last updated on 14/May/20

$$\mathrm{Find}\mathrm{the}\mathrm{n}-\mathrm{th}\mathrm{derivative}\mathrm{of}$$$$\mathrm{f}\left(x\right)=\sin \left(x\right)l\mathrm{n}x$$

Commented by mathmax by abdo last updated on 14/May/20

$$f^{\left(n\right)}\left(x\right)=\sum _{k=0}^n{C}_n^k{\left(lnx\right)}^{\left(k\right)}{\left(sinx\right)}^{(n-k)}$$$$=\left(lnx\right)sin(x+\frac{n\pi }{2})+\sum _{k=1}^n{C}_n^k{\left(lnx\right)}^{\left(k\right)}sin(x+\frac{(n-k)\pi }{2})$$$$wehave{\left(lnx\right)}^{{}^{\prime }}=\frac{1}{x}\Rightarrow {\left(ln\left(x\right)\right)}^{\left(k\right)}={\left(\frac{1}{x}\right)}^{(k-1)}=\frac{{(-1)}^{k-1}(k-1)!}{x^k}$$$$\Rightarrow f^{\left(n\right)}\left(x\right)=lnxsin(x+\frac{n\pi }{2})+\sum _{k=1}^n{C}_n^k\frac{{(-1)}^{k-1}(k-1)!}{x^k}\times sin(x+\frac{(n-k)\pi }{2})$$

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