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**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.