what is the nth derivative of f(x) if f(x) = square root of x?

Answer 1

f(x) = ax^n

f'(x) = (an)x^(n - 1)
f''(x) = (an(n - 1))x^((n - 1) - 1) = an(n - 1)x^(n - 2)
f'''(x) = an((n - 1)(n - 2))x^(n - 3)

so

nth derivative = a(n!)x^(n - (nth derivative))

in otherwords

f(x) = x^(1/2)

f'(x) = (1/2)x^((1/2) - 1) = (1/2)x^(-1/2)
f''(x) = (-1/4)x^((-1/2) - 1) = (-1/4)x^(-3/2)
f'''(x) = (3/8)x^(-5/2)
f''''(x) = (-15/16)x^(-7/2)

the single "n" value is what your value is being raised by
the nth derivative is just the nth derivative
"a" stands for the coefficient.

Answer 2

nth equals= (1/2)^n * (x)^(1/2-n)