Question

1) (x+1)√x
2) (2x-1)√x
3) x-1/√x
4) √x/2x+1
Производная должна быть рассчитана

Answer 1

Ответ:

1.

$y = (x + 1) \sqrt{x} = x \sqrt{x} + \sqrt{x} = \\ = {x}^{ \frac{3}{2} } + {x}^{ \frac{1}{2} }$

$y' = \frac{3}{2} {x}^{ \frac{1}{2} } + \frac{1}{2} {x}^{ - \frac{1}{2} } = \frac{3}{2} \sqrt{x} + \frac{1}{2 \sqrt{x} }$

2.

$y = (2x - 1) \sqrt{x} = 2x \sqrt{x} - \sqrt{x} = \\ = 2 {x}^{ \frac{3}{2} } - {x}^{ \frac{1}{2} }$

$y '= 2 \times \frac{3}{2} {x}^{ \frac{1}{2} } - \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 3 \sqrt{x} - \frac{1}{2 \sqrt{x} }$

3.

$y = \frac{ x - 1}{ \sqrt{x} } = \frac{x}{ \sqrt{x} } - \frac{1}{ \sqrt{x} } = \\ = \sqrt{x} - \frac{1}{ \sqrt{x} } = {x}^{ \frac{1}{2} } - {x}^{ - \frac{1}{2} }$

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

4.

$y = \frac{ \sqrt{x} }{2x + 1}$

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