elements of calculus

find ds/dt if s= sqaureroot of t over (3t -1)

asked by tara
  1. Using the quotient rule:

    s = √t / (3t-1)

    s' = [1/(2√t) * (3t-1) - √t(3)]/(3t-1)^2
    = [(3t-1)/(2√t) - 3√t]/(3t-2)^2
    = [3t-1 - 6t]/[2√t (3t-2)^2]
    = -(3t+1)/[2√t (3t-2)^2]

    or, using the product rule:

    s = √t * (3t-1)^-1

    s' = 1/(2√t) * (3t-1)^-1 - 3√t * (3t-1)^-2

    posted by Steve

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