Question

Заранеее огромное спасибо и 30 баллов за решение.

Answer 1

Ответ:

$1)f'(x) = ln(5) \times {5}^{x} + 3 + \frac{7}{2} {x}^{6} - \frac{1}{x}$

$2)f'(x) = 2 {x}^{4} - \frac{5}{ ln(5) \times x} - \frac{2}{ { \cos}^{2}x } - \frac{ {x}^{2} }{2} - \frac{4}{ \sqrt{1 - {x}^{2} } }$

$3)y '= \frac{1}{5} {z}^{ - \frac{4}{5} } = \frac{1}{5 \sqrt[5]{ {z}^{4} } }$

$4)y' = \frac{1}{2} \times ( - \frac{2}{3} ) {x}^{ - \frac{5}{3} } = \\ = - \frac{1}{3x \sqrt[3]{ {x}^{2} } }$

$5)y' = - \sin(x) \times ( \frac{27}{x} {x}^{4} - x + 2) \times (27 {x}^{3} - 1) \times \cos(x)$

$6)y' = \frac{1}{2 \sqrt{x} } ( {x}^{2} + \sin(x)) + (2x + \cos(x) ) \times ( \sqrt{x} - 3)$

$7)y' = \frac{ {e}^{x} (3 {x}^{3} - 7) - 9 {x}^{2} {e}^{x} }{ {(3 {x}^{3} - 7) }^{2} } = \\ = \frac{ {e}^{x}(3 {x}^{3} - 9 {x}^{2} - 7)}{ {(3 {x}^{3} - 7)}^{2} }$

$8)y' = \frac{4( {x}^{2} + 4) - 2x(4x - 7)}{ {( {x}^{2} + 4)}^{2} } = \\ = \frac{4 {x}^{2} + 16 - 8 {x}^{2} + 14x}{ {( {x}^{2} + 4)}^{2} } = \\ = \frac{ - 4 {x}^{2} + 14x + 16 }{ {( {x}^{2} + 4) }^{2} }$