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If f (2) = 1, \ f '(2) = - 3 and g (x) = log_2...

If f (2) = 1, \ f '(2) = - 3 and g (x) = log_2 (x ^ 2f (x) / cos (f (x)) then g' (2) is equal to:
a)-9.625983

B)-3.327777

c)-4.800967

d)-12.01444

E)The answer doesn´t exist

Math   Calculus   
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