Question

Найти производную функции

Answer 1

Ответ:

a

$f'(x) = \frac{1}{4} \times 4 {x}^{3} - \frac{1}{3} \times 3 {x}^{2} + \frac{1}{2} \times 2x - 1 + 0 = \\ = {x}^{3} - {x}^{2} + x - 1$

б

$f'(x) = (3 {x}^{ - 3} - {x}^{ \frac{1}{5} } + 5 {x}^{ - \frac{1}{3} } ) '= \\ - 9 {x}^{ - 4} - \frac{1}{5} {x}^{ - \frac{4}{5} } - \frac{5}{3} {x}^{ - \frac{4}{3} } = - \frac{9}{ {x}^{4} } - \frac{1}{5 \sqrt[5]{ {x}^{4} } } - \frac{5}{3x \sqrt[3]{x} }$

в

$f'(x) = \cos(3x) \times (3x)' - \sin(5x) \times (5x) '= \\ = 3 \cos(3x) - 5 \sin(5x)$

г

$f'(x) = ( {( {x}^{3} - 2 \cos(5x + 1)) }^{ \frac{1}{2} } ) '= \\ = \frac{1}{2} {( {x}^{3} - 2 \cos(5x + 1)) }^{ - \frac{1}{2} } \times ( {x}^{3} - 2 \cos(5x + 1)) '= \\ = \frac{1}{2 \sqrt{ {x}^{3} - 2 \cos(5x + 1) } } \times (3 {x}^{2} + 2 \sin(5x + 1) \times 5) = \\ = \frac{3 {x}^{2} + 10 \sin(5x + 1) }{ \sqrt{ {x}^{3} - 2 \cos(5x + 1) } }$