Question

Найти производные функции помогите решить​

Answer 1

Ответ:

а)

$y' = 0.5 \times 4 {x}^{3} - 0.2 \times ( - 5) {x}^{ - 6} = \\ = 2 {x}^{3} + \frac{1}{ {x}^{6} }$

б)

$y '= ( \sqrt{x} (2 {x}^{3} - 4x + 5))' = (2 {x}^{3} \sqrt{x} - 4x \sqrt{x} + 5 \sqrt{x} ) '= \\ = (2 {x}^{ \frac{7}{2} } - 4 {x}^{ \frac{3}{2} } + 5 {x}^{ \frac{1}{2} } ) '= 2 \times \frac{7}{2} {x}^{ \frac{5}{2} } - 4 \times \frac{3}{2} {x}^{ \frac{1}{2} } + 5 \times \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 7 {x}^{2} \sqrt{x} - 6 \sqrt{x} + \frac{5}{2 \sqrt{x} }$

в)

$y' = \frac{(x - 3 {x}^{2} )'(2 + x) - (x + 2)'(x - 3 {x}^{2} )}{ {(x + 2)}^{2} } = \\ = \frac{(1 - 6x)(x + 2) - (x - 3 {x}^{2}) }{ {(x + 2)}^{2} } = \\ = \frac{x + 2 - 6 {x}^{2} - 12x - x + 3 {x}^{2} }{ {(x + 2) }^{2} } = \frac{ - 3 {x}^{2} - 12x + 2}{ {(x + 2)}^{2} }$

г)

$y '= 5 {(1 + {e}^{x} )}^{4} \times (1 + {e}^{x} ) '= 5 {e}^{x} (1 + e {}^{x}) {}^{4}$