what-is-the-nth-derivative-of-e-ax-sin-bx-c-

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

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)