Question

найти производную функции

Answer 1

5.

$y '= \frac{u'v - v'u}{ {v}^{2} } = \\ = \frac{ - \sin( x ) \times ( \cos(x) - 1) - ( - \sin(x)) \times (1 + \cos(x)) }{ {( \cos(x) - 1)}^{2} } = \\ = \frac{ - \sin(x) \cos(x) + \sin(x) + \sin(x) + \sin(x) \cos(x) }{ {( \cos(x) - 1) }^{2} } = \\ = \frac{2 \sin(x) }{ {( \cos(x) - 1)}^{2} }$

$y'(60°) = \frac{2 \sin(60°) }{ {( \cos(60°) - 1)}^{2} } = \frac{2 \times \frac{ \sqrt{3} }{2} }{( \frac{1}{2} - 1) ^{2} } = \\ = \sqrt{3} \times \frac{1}{ \frac{1}{4} } = 4 \sqrt{3}$

6.

$f'(x) = \frac{1}{5} \times 5 {x}^{4} + \frac{10}{3} \times 3 {x}^{2} - 9 = \\ = {x}^{4} + 10 {x}^{2} - 9 \\ \\ {x}^{4} + 10 {x}^{2} - 9 = 0$

замена:

${x}^{2} = t \\ {t}^{2} + 10 t - 9 = 0 \\ D = 100 + 36 = 136 = 4 \times 34 \\ t1 = \frac{ - 10 + 2 \sqrt{34} }{2} = - 5 + \sqrt{34} \\ t2 = - 5 - \sqrt{34} \\ \\ {x}^{2} = - 5 + \sqrt{34} \\ x1 = \sqrt{ - 5 + \sqrt{34} } \\ x2 = - \sqrt{ - 5 + \sqrt{34} } \\ \\ {x}^{2} = - 5 - \sqrt{34}$

нет корней

Ответ:

$x1 = \sqrt{ \sqrt{34} - 5} \\ x2 = - \sqrt{ \sqrt{34} - 5}$