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

Answer 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