- #1
dnt
- 238
- 0
how do you find the general formula for the nth derivative of x^1/2 (square root of x)?
i can get the full formulas but i cannot put it together as a general formula:
n=1 (1st der): (1/2)(x^-1/2)
n=2 (2nd der): (1/2)(-1/2)(x^-3/2)
n=3 (3rd der): (1/2)(-1/2)(-3/2)(x^-5/2)
n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2)
...
nth der = ?
i have it right so far, correct? i just can't figure out how put it in terms of n (is it an arithmetic series or something)?
thanks.
i can get the full formulas but i cannot put it together as a general formula:
n=1 (1st der): (1/2)(x^-1/2)
n=2 (2nd der): (1/2)(-1/2)(x^-3/2)
n=3 (3rd der): (1/2)(-1/2)(-3/2)(x^-5/2)
n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2)
...
nth der = ?
i have it right so far, correct? i just can't figure out how put it in terms of n (is it an arithmetic series or something)?
thanks.