We will have to use product rule regardless, but it looks to be easier if we distribute the 3x to the binomial first:
h(x) = (3x3 + 3x)·(f(x))
h'(x) = (9x2 + 3)·f(x) + f'(x)·(3x3 + 3x)
h'(0) = 3·f(0) = 6
Michael R.
asked 10/01/21
We will have to use product rule regardless, but it looks to be easier if we distribute the 3x to the binomial first:
h(x) = (3x3 + 3x)·(f(x))
h'(x) = (9x2 + 3)·f(x) + f'(x)·(3x3 + 3x)
h'(0) = 3·f(0) = 6
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