elements of calculus
posted by tara
find ds/dt if s= sqaureroot of t over (3t 1)

Steve
Using the quotient rule:
s = √t / (3t1)
s' = [1/(2√t) * (3t1)  √t(3)]/(3t1)^2
= [(3t1)/(2√t)  3√t]/(3t2)^2
= [3t1  6t]/[2√t (3t2)^2]
= (3t+1)/[2√t (3t2)^2]
or, using the product rule:
s = √t * (3t1)^1
s' = 1/(2√t) * (3t1)^1  3√t * (3t1)^2
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