Questão 4 - Calculus
Find the derivative of the function f ( x ) = 2 x 3 − 4 x + 1 x 2 + 1 f(x) = \frac{2x^3 - 4x + 1}{x^2 + 1} Options: A) 6 x 2 ( x 2 + 1 ) − ( 2 x 3 − 4 x + 1 ) ( 2 x ) ( x 2 + 1 ) 2 \frac{6x^2(x^2 + 1) - (2x^3 - 4x + 1)(2x)}{(x^2 + 1)^2} B) ( 6 x 2 − 4 ) ( x 2 + 1 ) − ( 2 x 3 − 4 x + 1 ) ( 2 x ) ( x 2 + 1 ) 2 \frac{(6x^2 - 4)(x^2 + 1) - (2x^3 - 4x + 1)(2x)}{(x^2 + 1)^2} C) ( 6 x 2 − 4 ) ( x 2 + 1 ) − ( 2 x 3 − 4 x + 1 ) ( 2 ) ( x 2 + 1 ) 2 \frac{(6x^2 - 4)(x^2 + 1) - (2x^3 - 4x + 1)(2)}{(x^2 + 1)^2} D) ( 6 x 2 − 4 ) ( x 2 + 1 ) − ( 2 x 3 − 4 x + 1 ) ( x ) ( x 2 + 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