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Question Number 47432 by Necxx last updated on 09/Nov/18

what is the nth derivative of e^(ax) sin (bx+c)?

$$ \\ $$$${what}\:{is}\:{the}\:{nth}\:{derivative}\:{of} \\ $$$${e}^{{ax}} \mathrm{sin}\:\left({bx}+{c}\right)? \\ $$

Commented by maxmathsup by imad last updated on 09/Nov/18

let f(x)=e^(ax) sin(bx+c) ⇒f(x)=Im(e^(ax) e^(i(bx+c)) ) =Im( e^(ic) e^((a+ib)x) ) ⇒f^((n)) (x)=Im((e^(ic) e^((a+ib)x) )^((n)) ) but {e^(ic) e^((a+ib)x) }^((n)) =e^(ic) { e^((a+ib)x) }^((n)) =e^(ic) (a+ib)^n e^((a+ib)x) =e^(ax) e^(i(bx+c)) r^n e^(inθ) with r=(√(a^2 +b^2 )) cosθ =(a/(√(a^2 +b^2 ))) and sinθ =(b/(√(a^2 +b^2 )))(⇒ tanθ=(b/a) ⇒θ =arctan((b/a)) ⇒ e^(ax) e^(i(bx+c)) r^n e^(inθ) =r^n e^(ax) e^(i(bx+c +nθ)) =r^n e^(ax) {cos(nθ +bx+c)+i sin(nθ +bx+c)} ⇒ f^((n)) (x)= r^n e^(ax) sin(nθ +bx+c) with r=(√(a^2 +b^2 )) and θ =arctan((b/a)) ( a≠0)

$${let}\:{f}\left({x}\right)={e}^{{ax}} {sin}\left({bx}+{c}\right)\:\Rightarrow{f}\left({x}\right)={Im}\left({e}^{{ax}} \:{e}^{{i}\left({bx}+{c}\right)} \right) \\ $$$$={Im}\left(\:{e}^{{ic}} \:{e}^{\left({a}+{ib}\right){x}} \right)\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)={Im}\left(\left({e}^{{ic}} \:{e}^{\left({a}+{ib}\right){x}} \right)^{\left({n}\right)} \right)\:{but} \\ $$$$\left\{{e}^{{ic}} \:{e}^{\left({a}+{ib}\right){x}} \right\}^{\left({n}\right)} ={e}^{{ic}} \:\left\{\:{e}^{\left({a}+{ib}\right){x}} \right\}^{\left({n}\right)} \:={e}^{{ic}} \left({a}+{ib}\right)^{{n}} \:{e}^{\left({a}+{ib}\right){x}} \\ $$$$={e}^{{ax}} \:{e}^{{i}\left({bx}+{c}\right)} \:{r}^{{n}} \:{e}^{{in}\theta} \:\:\:{with}\:{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:{cos}\theta\:\:=\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:{and}\:{sin}\theta\:=\frac{{b}}{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\left(\Rightarrow\right. \\ $$$${tan}\theta=\frac{{b}}{{a}}\:\Rightarrow\theta\:={arctan}\left(\frac{{b}}{{a}}\right)\:\Rightarrow \\ $$$${e}^{{ax}} \:{e}^{{i}\left({bx}+{c}\right)} \:{r}^{{n}} \:{e}^{{in}\theta} \:={r}^{{n}} \:{e}^{{ax}} \:{e}^{{i}\left({bx}+{c}\:+{n}\theta\right)} \\ $$$$={r}^{{n}} {e}^{{ax}} \left\{{cos}\left({n}\theta\:+{bx}+{c}\right)+{i}\:{sin}\left({n}\theta\:+{bx}+{c}\right)\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\:{r}^{{n}} \:{e}^{{ax}} \:{sin}\left({n}\theta\:+{bx}+{c}\right)\:{with}\:{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{and}\:\theta\:={arctan}\left(\frac{{b}}{{a}}\right)\:\:\left(\:{a}\neq\mathrm{0}\right) \\ $$

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