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

Find the n-th derivative of f(x)=sin(x)lnx

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

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

f^((n)) (x) =Σ_(k=0) ^n C_n ^k (lnx)^((k)) (sinx)^((n−k)) =(lnx)sin(x+((nπ)/2)) +Σ_(k=1) ^n C_n ^k (lnx)^((k)) sin(x+(((n−k)π)/2)) we have (lnx)^′ =(1/x) ⇒(ln(x))^((k)) =((1/x))^((k−1)) =(((−1)^(k−1) (k−1)!)/x^k ) ⇒f^((n)) (x) =lnx sin(x+((nπ)/2))+Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/x^k )×sin(x+(((n−k)π)/2))

$${f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({lnx}\right)^{\left({k}\right)} \:\left({sinx}\right)^{\left({n}−{k}\right)} \\ $$$$=\left({lnx}\right){sin}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({lnx}\right)^{\left({k}\right)} {sin}\left({x}+\frac{\left({n}−{k}\right)\pi}{\mathrm{2}}\right) \\ $$$${we}\:{have}\:\left({lnx}\right)^{'} \:=\frac{\mathrm{1}}{{x}}\:\Rightarrow\left({ln}\left({x}\right)\right)^{\left({k}\right)} \:=\left(\frac{\mathrm{1}}{{x}}\right)^{\left({k}−\mathrm{1}\right)} \:=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{{x}^{{k}} } \\ $$$$\Rightarrow{f}^{\left({n}\right)} \left({x}\right)\:={lnx}\:{sin}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{{x}^{{k}} }×{sin}\left({x}+\frac{\left({n}−{k}\right)\pi}{\mathrm{2}}\right) \\ $$

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