Respostas

2013-10-04T15:28:22-03:00
f(x) = \frac{18x^{2}+6x-3x-1}{x}

f(x) = \frac{18x^{2}+3x-1}{x}

derivada do quociente

(\frac{f(x)}{g(x)})'= \frac{f'(x)* g(x)- (f(x)* g'(x))}{(g(x))^{2}}

ENTÃO:
f(x) = ( \frac{3x +1}{x}) * (6x - 1)

f(x) =  \frac{18x^{2} + 6x - 3x - 1}{x}

f(x) = \frac{18x^{2} + 3x - 1}{x}

f'(x) = \frac{(36x + 3)*x - ((18x^{2}+3x-1)* 1)}{x^{2}}

f'(x) = \frac{36x^{2} + 3x - 18x^{2}-3x+1}{x^{2}}

RESPOSTA
f'(x) = \frac{18x^{2} +1}{x^{2}}