Answer :
Here,
f(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8)
f(1) = (2)(2)(2)(2) = 16
Taking log on both sides we get,
Log (f (x)) = log (1 + x) + log (1 + x2) + log (1 + x4) + log(1 + x8)
Differentiating it with respect to x we get,
F’(1) = 120
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