# formula for the n:th derivative

• Oct 9th 2010, 01:03 PM
Zamzen
formula for the n:th derivative
Find the formula for the n:th derivative y^(n) for the function y=1/x^(1/2)'

I started by finding out the formula for the functions up to the fourth derivative.
and i came as far as
y = y=1/x^(1/2)
y' = y= -1/2x^(3/2)
y'' =3/4x^(5/2)
y'''=-15/8x^(1/2)
y''''=105/16x^(9/2)

And the pattern which i found is that y^n where n starts from 0. so 0 is the original function.
The nth derivative y^(n) = (-1)^n / ((2^n)*x^(0.5 +n)). The problem here is that i dont find anything that can fix the numerator. I realise the pattern is 1*3*5*7...n. But i dont know how to write a value for this since i cant use ! because its 1*2*3*4 etc. Is it possible to write a factorial equation that skipps whole numbers and only takes odd numbers? And if so, how do i write it mathematically?
Any help is appreciated.
• Oct 9th 2010, 01:21 PM
yeKciM
Quote:

Originally Posted by Zamzen
Find the formula for the n:th derivative y^(n) for the function y=1/x^(1/2)'

I started by finding out the formula for the functions up to the fourth derivative.
and i came as far as
y = y=1/x^(1/2)
y' = y= -1/2x^(3/2)
y'' =3/4x^(5/2)
y'''=-15/8x^(1/2)
y''''=105/16x^(9/2)

And the pattern which i found is that y^n where n starts from 0. so 0 is the original function.
The nth derivative y^(n) = (-1)^n / ((2^n)*x^(0.5 +n)). The problem here is that i dont find anything that can fix the top. I realise the pattern is 1*3*5*7...n. But i dont know how to write a value for this since i cant use ! because its 1*2*3*4 etc. Is it possible to write a factorial equation that skipps whole numbers and only takes odd numbers? And if so, how do i write it mathematically?
Any help is appreciated.

(2n)!! = 2n * (2n-2) *(2n-4) * .... == even numbers :D
(2n+1)!! = (2n+1)*(2n-1) *(2n-3) * ..... == odd numbers :D
• Oct 9th 2010, 02:15 PM
undefined
Quote:

Originally Posted by yeKciM
(2n)!! = 2n * (2n-2) *(2n-4) * .... == even numbers :D
(2n+1)!! = (2n+1)*(2n-1) *(2n-3) * ..... == odd numbers :D

$\displaystyle 2(4)(6)(8)\cdot\cdot\cdot (2(n-1))(2n)= 2^n(1(2)(3)(4)\cdot\cdot\cdot (n-1)(n))= 2^n n!$.
$\displaystyle 1(3)(5)(7)\cdot\cdot\cdot (2n-1)(2n+1)= \frac{1(2)(3)(4)(5)(6)(7)\cdot\cdot\cdot\(2n-2)(2n)(2n+1)}{2(4)(6)\cdot\cdot\cdot (2n-2)(2n)}= \frac{(2n+1)!}{2^n n!}$