Question

nth derivative of y=e^ax*cos(bx)

Answer 1

i got it no tension

Answer 2

i think u write it in the form\[y=e^{ax}\times\cos(bx)=e^{ax}\times \frac{e^{ibx}+e^{-ibx}}{2}\]

Answer 3

but how can cosbx be written as what u have written

Answer 4

are u familiar with complex numbers?

Answer 5

i have solved by substituting \[a=k \sin \alpha\]
\[b=k \sin \alpha\]

Answer 6

not much familiar with them please explain iam in 10th jst doing advanced studies

Answer 7

@TuringTest lets work on this

Answer 8

what was that formula u used for a problem once ?

Answer 9

Leibniz theorem...

Answer 10

\[(uv)^n ...\]

Answer 11

\[(uv)'=\sum_{k=0}^n\binom nku^{(k)}v^{(n-k)}\]

Answer 12

looks just like the binomial theorem

Answer 13

unfortunately I actually can't focus on this too much right now, but I will see if I find something fairly obvious...

Answer 14

that should be (uv)^(n) like you said...

Answer 15

np :) thank u

Answer 16

\[(uv)^{(n)}=\sum_{k=0}^n\binom nku^{(k)}v^{(n-k)}\]

Answer 17

ok i'll start letting\[u=\cos bx\]\[v=e^{ax}\]

Answer 18

that formula works better usually when it says "find the nth derivative at x=0 or something, but it may help us here"

Answer 19

..and I have to go, but I will be happy to see how this works out :)
see ya!

Answer 20

cya :)

Answer 21

\[(\cos bx .e^{ax})^{(n)}=\sum_{k=0}^n\binom nk(\cos bx)^{(k)}(e^{ax})^{(n-k)}\]

Answer 22

its immidiate that\[(e^{ax})^{(n-k)}=a^{n-k} e^{ax}\]

Answer 23

\[(\cos bx)^{(k)}=b^k \cos(\frac{k\pi}{2}+bx)\]

Answer 24

i think i'll work on this later...gotta go to bed :)

Answer 25

i think it wud be difficult to do it like that by using lebinitz theorem

Answer 26

yeah it is compilicated

Answer 27

the solution is possible by substituting \[a=k \cos \alpha\]
\[b=k \sin \alpha\]

Answer 28

why letting\[a=b=k \sin \alpha\]????

Answer 29

no not that because that compulsorily means that u r making a and b equal but with one expressed as product of \[ \cos \alpha\] and \[\sin \alpha\] you have exactly a real situation