Differentiate.

g(x)=(x^3+1)(3x^2-1)

I know that the answer is
15x^4-3x^2+6x but I donot know how to get to this. I was using
g'(x) X h(x) + g(x) X h'(x) and I was using the derivatives 3x^2 and 6x but I guess this is wrong. Please help!! Thank you.

2 answers

g '(x) = 6x(x^3+1) + 3x^2(3x^2-1)
= 6x^4 + 6x + 9x^4 - 3x^2
= 15x^4 - 3x^2 + 6x
Thank You
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