Christopher G. answered • 11/29/13

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f(x) = x

^{2/3}(2 - x)I would first distribute to make it easier to derive (otherwise we would have to use the product rule and that can get messy).

f(x) = 2x

^{2/3}- x^{5}^{/3}(remember x is really x^{1}or x^{3/3}in this case, so x^{2/3}times x^{3/3}is x^{5/3})Now use the power rule:

f'(x) = 2(2/3)x

^{-1/3}- (5/3)x^{2/3}= (4/3)x^{-1/3}- (5/3)x^{2/3}Now do the same thing to f'(x) to get f"(x):

f"(x) = (4/3)(-1/3)x

^{-4/3}- (5/3)(2/3)x^{-1/3}= (-4/9)x^{-4/3}- (10/9)x^{-1/3}