Question

Вычислите значение производной функции в точке x0= - 1
$y=\frac{(x-1)^{3} }{x^{2} +1}$

Answer 1

Ответ:

$y = \frac{ {(x - 1)}^{3} }{ {x}^{2} + 1 }$

$y' = \frac{( {(x - 1)}^{3}) '( {x}^{2} + 1) - ( {x}^{2} + 1)' {(x - 1)}^{3} }{ {( {x}^{2} + 1) }^{2} } = \\ = \frac{3 {(x - 1)}^{2} ( {x}^{2} + 1) - 2x {(x - 1)}^{3} }{ {( {x}^{2} + 1)}^{2} } = \\ = \frac{ {(x - 1)}^{2} (3( {x}^{2} + 1) - 2x(x - 1)) }{ {( {x}^{2} + 1)}^{2} } = \\ = \frac{ {(x - 1)}^{2}(3 {x}^{2} + 3 - 2 {x}^{2} + 2x) }{ {( {x}^{2} + 1)}^{2} } = \\ = \frac{ {(x - 1)}^{2}( {x}^{2} + 2x + 3) }{ {( {x}^{2} + 1)}^{2} }$

$y'( - 1) = \frac{4 \times (1 - 2 + 3)}{4} = 2$