How do you integrate #y=(2x^3)/(4-x)# using the quotient rule?

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Answer 1 · 2016-09-16T12:32:51

#(4 x^(2) (6 - x)) / ((4 - x)^(2))#

The question should be asking how to differentiate the given function, and not how to integrate it.

We have: #y = (2 x^(3)) / (4 - x)#

This function can be differentiated using the "quotient rule":

#=> y' = ((4 - x) (6 x^(2)) - (2 x^(3)) (- 1)) / ((4 - x)^(2))#

#=> y' = (24 x^(2) - 6 x^(3) + 2 x^(3)) / ((4 - x)^(2))#

#=> y' = (24 x^(2) - 4 x^(3)) / ((4 - x)^(2))#

#=> y' = (4 x^(2) (6 - x)) / ((4 - x)^(2))#