Question

is there any easy formula for the nth derivative of (t^a)e^(bt) w.r.t. time? (a and b are constants...)

Answer 1

You can definitely derive one, if you have the patience.
\[\large\begin{align*}f(t)&=t^ae^{bt}\\
f'(t)&=at^{a-1}e^{bt}+bt^ae^{bt}\\
&=\left(at^{a-1}+bt^a\right)e^{bt}\\
f''(t)&=\left(a(a-1)t^{a-2}+abt^{a-1}\right)e^{bt}+\left(bat^{a-1}+b^2t^a\right)e^{bt}\\
&=\left(a(a-1)t^{a-2}+2abt^{a-1}+b^2t^a\right)e^{bt}\\
f'''(t)&=\left(a(a-1)(a-2)t^{a-3}+2a(a-1)bt^{a-2}+ab^2t^{a-1}\right)e^{bt}\\&\;\;\;\;\;+\left(ba(a-1)t^{a-2}+2ab^2t^{a-1}+b^3t^a\right)e^{bt}\\
&=\left(a(a-1)(a-2)t^{a-3}+3ba(a-1)t^{a-2}+3ab^2t^{a-1}+b^3t^a\right)e^{bt}\\
&\vdots\\
f^{(n)}(t)&=\cdots\end{align*}\]
I can see the pattern, but I'm not totally sure how to write it out. The formula will involve an a! and a binomial expansion, as per the Binomial Theorem.

Answer 2

yes, but this is still a mess.. try manipulating a bit more.

Answer 3

Is my post cutoff for anyone else, or just me? I can only see half of it, and it's completely shifted upward.

Answer 4

try Leibniz rule

Answer 5

yeah .. that is similar to binomial rule.

Answer 6

do you try recurrence function?

Answer 7

what recurrence function? probably yes ... give it a try.

Answer 8

I just think about it because it looks like what I study recurrence function. I try to make it have the form of f(sub n+2) + f (sub n+1) +f (sub n) =0 with initial value of f (0) is the original one.

Answer 9

Just idea. I have no time to do. you do it, experimentX

Answer 10

this is your recurrence relation ... guess it doesn't help in finding the nth derivative.