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


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 47432 by Necxx last updated on 09/Nov/18

$w h a t i s t h e n t h d e r i v a t i v e o f$ $e^{a x} \sin (b x + c) ?$
Commented by maxmathsup by imad last updated on 09/Nov/18
$let f(x)=e^(ax) sin(bx+c) ⇒f(x)=Im(e^(ax) e^(i(bx+c)) ) =Im( e^(ic) e^((a+ib)x) ) ⇒f^((n)) (x)=Im((e^(ic) e^((a+ib)x) )^((n)) ) but {e^(ic) e^((a+ib)x) }^((n)) =e^(ic) { e^((a+ib)x) }^((n)) =e^(ic) (a+ib)^n e^((a+ib)x) =e^(ax) e^(i(bx+c)) r^n e^(inθ) with r=(√(a^2 +b^2 )) cosθ =(a/(√(a^2 +b^2 ))) and sinθ =(b/(√(a^2 +b^2 )))(⇒ tanθ=(b/a) ⇒θ =arctan((b/a)) ⇒ e^(ax) e^(i(bx+c)) r^n e^(inθ) =r^n e^(ax) e^(i(bx+c +nθ)) =r^n e^(ax) {cos(nθ +bx+c)+i sin(nθ +bx+c)} ⇒ f^((n)) (x)= r^n e^(ax) sin(nθ +bx+c) with r=(√(a^2 +b^2 )) and θ =arctan((b/a)) ( a≠0)$
$l e t f (x) = e^{a x} s i n (b x + c) \Rightarrow f (x) = I m (e^{a x} e^{i (b x + c)})$ $= I m (e^{i c} e^{(a + i b) x}) \Rightarrow f^{(n)} (x) = I m ({(e^{i c} e^{(a + i b) x})}^{(n)}) b u t$ ${e^{i c} e^{(a + i b) x}}^{(n)} = e^{i c} {e^{(a + i b) x}}^{(n)} = e^{i c} {(a + i b)}^{n} e^{(a + i b) x}$ $= e^{a x} e^{i (b x + c)} r^{n} e^{i n θ} w i t h r = \sqrt{a^{2} + b^{2}} c o s θ = \frac{a}{\sqrt{a^{2} + b^{2}}} a n d s i n θ = \frac{b}{\sqrt{a^{2} + b^{2}}} (\Rightarrow$ $t a n θ = \frac{b}{a} \Rightarrow θ = a r c t a n (\frac{b}{a}) \Rightarrow$ $e^{a x} e^{i (b x + c)} r^{n} e^{i n θ} = r^{n} e^{a x} e^{i (b x + c + n θ)}$ $= r^{n} e^{a x} {c o s (n θ + b x + c) + i s i n (n θ + b x + c)} \Rightarrow$ $f^{(n)} (x) = r^{n} e^{a x} s i n (n θ + b x + c) w i t h r = \sqrt{a^{2} + b^{2}} a n d θ = a r c t a n (\frac{b}{a}) (a \neq 0)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum