How do you differentiate #y= 12(x^2-7)^(1/3)#?

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Answer 1 · 2015-04-25T19:40:10

I would use the Chain Rule to deal with the #()^(1/3)#:
#y'=12*1/3(x^2-7)^(1/3-1)*2x=8x(x^2-7)^(-2/3)#