Asked by tara
find ds/dt if s= sqaureroot of t over (3t -1)
Answers
Answered by
Steve
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
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
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