So, I assume that you mean taking the n-th derivative of f(x) for f(x) = x(-3/2). If that is the case then thinking about it in the same way we would think about a single derivative f'(x) = (-3x(-5/2))/2.
We can then look at f''(x) = (-3)*(-5)*x(-7/2))/(2*2).
So if you know what the factorial function is, there is something really similar to that, that we can use.
So if you do not know what factorial is then here: k! = k*(k - 1)*(k - 2)*...*2*1 which grows extremely fast, eventually faster than the exponential function.
However even though (k!)! grows much faster than k! since we are doing the operation twice, there is a different type of operation called "double factorial" which is not the same as taking the factorial twice.
The double factorial is as follows k!! = k*(k - 2)*(k - 4)*...(m + 2)*m
where I put an m and (m + 2) instead of putting the last two terms in relation to k or being constants because the double factorial has two "parities", which means that even though it is defined the same way, we get different results whether k is even or odd. So if k is odd or in other words k = 2j + 1 (definition of odd numbers) then every term k, (k - 2), (k - 4), ... will be odd which means m = 1, (m + 2) = 3, ... all the way up until k. Whereas if k is even or in other words k = 2j then every term k, (k - 2), (k - 4), ... will be odd which means m = 2, (m + 2) = 4, ... all the way up until k.
With all that being said, there are some very complicated ways to transform double factorial into terms of single factorial, but I wouldn't recommend worrying to long about that. The formula you are looking for is the following:
the n-th derivative of f(x) is ((-1/2)n((2n + 1)!!)x(-(2n + 3)/2))
We know this since as n increases by 1 the term we multiply becomes -(2n + 1)/2 and we decrease the power by 1, which is the same as adding -2/2 to the power.
I hope this helped! If you have any other questions reply to this comment or book a lesson with me.
Ethan Bartiromo
