Asked by MeGaTrOn
y + ((4x^2)/(3-x)) ^3
y ' = ?
Product Rule didn't work here ...I think the Quotient rule but the cube on end has me stumped
y ' = ?
Product Rule didn't work here ...I think the Quotient rule but the cube on end has me stumped
Answers
Answered by
Steve
If you let
u = 4x^2
v = 3-x
then you have
y = (u/v)^3
y' = 3(u/v)^2 (u/v)'
= 3(u/v)^2 ((u'v-uv')/v^2)
= 192x^5(6-x) / (3-x)^4
u = 4x^2
v = 3-x
then you have
y = (u/v)^3
y' = 3(u/v)^2 (u/v)'
= 3(u/v)^2 ((u'v-uv')/v^2)
= 192x^5(6-x) / (3-x)^4
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