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 cant figure out how put it in terms of n (is it an arithmetic series or something)? thanks.