If f(x) = {(3+x)/(1+x)}^2+3x , find f '(0).

BUT IT'S ANS IS (27LOG3-12)

i made a mistake, i omitted some part while using product rule.

f(x) = {(3+x)/(1+x)}^2+3x

taking log on both sides

log(fx) = (2 + 3x) log {(3+x)/(1+x)}

log(fx) = (2 + 3x) [log(3+x) - log(1+x)]

Now differentiate using product rule

f'(x) /f(x) = (2 + 3x) [1/(3+x) - 1/(1 + x)] + [log(3+x) - log(1+x)] x 3

f'(x) /f(x) = (2 + 3x) [1/(3+x) - 1/(1 + x)] + log {(3+x)/(1+x)} x 3

f'(x) /f(x) = (2 + 3x) [-2 / (3+x)(1+x)] + 3 log {(3+x)/(1+x)}

f'(x) = f(x) [(2 + 3x) [-2 / (3+x)(1+x)] + 3 log {(3+x)/(1+x)}]

f'(0) = f(0) [(2) [- 2 / 3] + 3 log (3/1)]

f'(0) = f(0) [(-4 / 3) + 3log3]

f(0) = {(3+0)/(1+0)}² = 9

f'(0) = 9 x [-4 /3 + 3log3]

f'(0) = -12 + 27log3

f'(0) = 27log3 - 12