Question

Срочно!!! Знатоки, найдите производную графика функции y=f(x)​

Answer 1

Ответ:

$a)f'(x) = ( {x}^{2} \sqrt{x} + 2x \sqrt{x} )' = \\ = ( {x}^{ \frac{5}{2} } + 2 {x}^{ \frac{3}{2} } )' = \\ = \frac{5}{2} {x}^{ \frac{3}{2} } + 2 \times \frac{3}{2} {x}^{ \frac{1}{2} } = \\ = \frac{5}{2} x \sqrt{x} + 3 \sqrt{x}$

$d)f'(x) = 5 {x}^{4} \cos(x) - \sin(x) \times {x}^{5} = \\ = {x}^{4} (5 \cos(x) - x \sin(x))$

$g)f'(x) = 1 \times log_{2}(x) + x \times \frac{1}{ ln(2) \times x} = \\ = log_{2}(x) + \frac{1}{ ln(2) }$

$j)f'(x) = \frac{2x(3x + 1) - 3 {x}^{2} }{ {(3x + 1)}^{2} } = \\ = \frac{6 {x}^{2} + 2x - 3 {x}^{2} }{ {(3x + 1)}^{2} } = \\ = \frac{3 {x}^{2} + 2x}{ {(3x + 1)}^{2} }$

$m)f'(x) = \frac{ {e}^{x}( {e}^{x} + 1) - {e}^{x} ( {e}^{x} - 1) }{ {( {e}^{x} + 1)}^{2} } = \\ = \frac{ {e}^{2x} + {e}^{x} - {e}^{2x} + {e}^{x} }{ {( {e}^{x} + 1) }^{2} } = \\ = \frac{2 {e}^{x} }{ {( {e}^{x} + 1) }^{2} }$