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Question

The derivative of fx=x2x31loge t dt, x>0, is

(a) 13 ln x

(b) 13 ln x-12 ln x

(c) (ln x)−1 x (x − 1)

(d) 3x2ln x

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Solution

(c) (ln x)−1 x (x − 1)

Using Newton Leibnitz formula

f'(x)=1logex3(3x2)1logex2(2x)=3x23lnx2x2lnx=x2lnxxlnx=1lnxx(x1)=(lnx)1x(x1)

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