Questão 4 - Calculus

 Find the derivative of the function 

f(x)=2x3−4x+1x2+1f(x) = \frac{2x^3 - 4x + 1}{x^2 + 1}

Options:

A) 6x2(x2+1)−(2x3−4x+1)(2x)(x2+1)2\frac{6x^2(x^2 + 1) - (2x^3 - 4x + 1)(2x)}{(x^2 + 1)^2}
B) (6x2−4)(x2+1)−(2x3−4x+1)(2x)(x2+1)2\frac{(6x^2 - 4)(x^2 + 1) - (2x^3 - 4x + 1)(2x)}{(x^2 + 1)^2}
C) (6x2−4)(x2+1)−(2x3−4x+1)(2)(x2+1)2\frac{(6x^2 - 4)(x^2 + 1) - (2x^3 - 4x + 1)(2)}{(x^2 + 1)^2}
D) (6x2−4)(x2+1)−(2x3−4x+1)(x)(x2+1)2\frac{(6x^2 - 4)(x^2 + 1) - (2x^3 - 4x + 1)(x)}{(x^2 + 1)^2}
E) None of the above​

Original idea by: João Augusto Ferreira de Moura

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