Question

Помогите найти производную функции

Answer 1

Ответ:

1.

1)

$y' = (12 {x}^{2} - {x}^{ \frac{1}{2} } ) '= 12 \times 2x - \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 24x - \frac{1}{2 \sqrt{x} }$

2)

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

3)

$y '= ( {x}^{3} + 4 {x}^{2} - 5{x}^{ - 2} ) '= 3 {x}^{2} + 8x + 10 {x}^{ - 3} = \\ = 3 {x}^{2} + 8x + \frac{10}{ {x}^{3} }$

4)

$y' = (\frac{ {x}^{5} + 4 {x}^{4} - 1}{ {x}^{2} }) '= ( {x}^{3} + 4 {x}^{2} - \frac{1}{ {x}^{2} } ) '= \\ = ( {x}^{3} + 4 {x}^{2} - {x}^{ - 2} ) '= 3 {x}^{2} + 8x + 2 {x}^{ - 3} = \\ = 3 {x}^{2} + 8x + \frac{2}{ {x}^{3} }$

5)

$y' = x '\sin(x) + ( \sin(x) )' \times x = \\ = \sin(x) + x \cos(x)$

6)

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

7)

$y' = - \sin(2x + \frac{\pi}{3} ) \times (2x + \frac{\pi}{3} ) '= \\ = - 2 \sin(2x + \frac{\pi}{3} )$

2.

$f(x) = x - \cos(x) \\ f'(x) = 1 + \sin(x) \\ \\ f'(x) = 0 \\ 1 + \sin(x) = 0 \\ \sin(x) = - 1 \\ x = - \frac{\pi}{2} + 2\pi \: n$

n принадлежит Z.