Question

Найти производные функции.

Answer 1

$1)y' = \frac{ {x}^{2} - 2x(x - 4) }{ {x}^{4} } = \\= \frac{ {x}^{2} - 2 {x}^{2} + 8x}{ {x}^{4} } = \frac{ - {x}^{2} + 8x}{ {x}^{4} } = \frac{8 - x}{ {x}^{3} }$

$2)y' = \frac{(5 {x}^{4} - 1)(x - 1) - ( {x}^{5} - x) }{ {(x - 1)}^{2} } = \\=\frac{5 {x}^{5} - 5 {x}^{4} - x + 1 - {x}^{5} + x}{ {(x - 1)}^{2} } = \\= \frac{4 {x}^{5} - 5 {x}^{4} + 1 }{ {(x - 1)}^{2} }$

$3)y' = \frac{ {e}^{x} \cos(x) + {e}^{x} \sin(x) }{ { \cos }^{2}(x) } = \\= \frac{ {e}^{x} ( \cos(x) + \sin(x)) }{ { \cos }^{2}(x) }$

$4)y' = \\= \frac{ \frac{1}{x} ( {x}^{2} - 2) - 2x ln(x) }{ {( {x}^{2} - 2)}^{2} } =\\= \frac{x - \frac{2}{x} - 2x ln(x) }{ {( {x}^{2} - 2) }^{2} }$

$5)y' = \frac{ {x}^{2} + 1 - 2x(x - 1)}{ {( {x}^{2} + 1) }^{2} } = \frac{ {x}^{2} + 1 - 2 {x}^{2} + 2x }{ {( {x}^{2} + 1) }^{2} } = \\= \frac{ - {x}^{2} + 2x + 1 }{ {( {x}^{2} + 1)}^{2} }$

$6)y' = \frac{ {x}^{2} + 8 - 2x \times x }{ {( {x}^{2} + 8)}^{2} } = \frac{ - {x}^{2} + 8 }{ {( {x}^{2} + 8) }^{2} }$