If f(x)= (x+1)^7 1+x^2 (x^2-x+1 )^6 then the value of f^ (0) is equal to

Step-by-step solution

Given f(x) = (x+1)^7 1+x^2 (x^2-x+1)^6 . Taking the natural logarithm on both sides: f(x) = 7 (x+1) + 1 2 (1+x^2) - 6 (x^2-x+1) . Differentiating both sides with respect to x : f'(x) f(x) = 7 x+1 + 1 2 2x 1+x^2 - 6 2x-1 x^2-x+1 . f'(x) f(x) = 7 x+1 + x 1+x^2 - 6(2x-1) x^2-x+1 . Evaluating at x=0 : f(0) = (0+1)^7 1+0^2 (0^2-0+1)^6 = 1 1 1 = 1 . Substituting x=0 into the derivative expression: f'(0) f(0) = 7 0+1 + 0 1+0^2 - 6(2(0)-1) 0^2-0+1 . f'(0) 1 = 7 + 0 - 6(-1) = 7 + 6 = 13 . Therefore, f'(0) = 13 . Answer: 13