How do you differentiate #u= 8/(3x^2)#?

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Answer 1 · 2015-04-15T22:47:42

#u = 8/(3x^2)#

could be written as

#u = 8/3 x^(-2)#

so

#(du)/(dx) = (-2)*8/3 x^(-3)#

#= - 16/(3x^3)#

Answer 2 · 2015-04-15T23:29:04

Apr 15, 2015

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