Question

derive
$y = \frac{x + 1}{ {x}^{2} + 3 }$
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Answer 1

Resposta:

Explicação passo-a-passo:

$Se \ y= \frac{f}{g} => y'= \frac{f'g-fg'}{g^2}\\\\y= \frac{x+1}{x^2+3} => \\f= x+1 => f'=1\\\\g=x^2+3=>g'=2x\\\\y'=\frac{1*(x^2+3)-(x+1)(2x)}{(x^2+3)^2} = \frac{x^2+3-(2x^2+2x)}{(x^2+3)^2}=\frac{-x^2-2x+3}{x^4+6x^2+9}$