If f(x)= ( 3+x 1+x )^ 2+3 x , then f^ (0)=

Step-by-step solution

Given f(x) = ( 3+x 1+x )^ 2+3x . Taking the natural logarithm on both sides: f(x) = (2+3x) ( 3+x 1+x ) = (2+3x) [ (3+x) - (1+x)] . Differentiating both sides with respect to x : 1 f(x) f'(x) = 3 [ (3+x) - (1+x)] + (2+3x) [ 1 3+x - 1 1+x ] . At x=0 , f(0) = ( 3 1 )^2 = 9 . Substituting x=0 into the derivative expression: 1 f(0) f'(0) = 3 [ 3 - 1] + (2) [ 1 3 - 1 1 ] . 1 9 f'(0) = 3 3 + 2 [ 1-3 3 ] = 3 3 + 2 (- 2 3 ) = 3 3 - 4 3 . f'(0) = 9 (3 3 - 4 3 ) = 27 3 - 12 . Answer: -12+27 3