How do you use the quotient rule to differentiate #y =((3t+4)/(6t+7))^3#?

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Answer 1 · 2016-07-17T16:20:51

#= -9(3t+4)^2/(6t+7)^4 #

#y =((3t+4)/(6t+7))^3#

a spot of chain rule first to set it up

#d/dt ((3t+4)/(6t+7))^3= 3((3t+4)/(6t+7))^2 d/dt ((3t+4)/(6t+7))#

and that's were the quotient rule comes in #(u/v)' = (u'v - uv')/v^2#

#d/dt ((3t+4)/(6t+7)) = (3*(6t +7) - (3t+4)*6)/(6t+7)^2 #

#= -3/(6t+7)^2 #

putting it all together

#d/dt ((3t+4)/(6t+7))^3= 3((3t+4)/(6t+7))^2 * -3/(6t+7)^2 #

#= -9(3t+4)^2/(6t+7)^4 #