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

Question

1446 views around the world

You can reuse this answer
Creative Commons License

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)#