Question

ДАЮ 50 БАЛІВ!
(АЛГЕБРА)

Answer 1

5)

$y = 4 {x}^{2} - {x}^{3} \\ y' = 4 \times 2 {x}^{2 - 1} - 3 {x}^{3 - 1} = 8x - 3 {x}^{2} \\ x_{o} = 1 \\ y '= 8 \times 1 - 3 \times {1}^{2} = 8 - 3 = 5$

6)

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

7)

$y = \frac{2x {}^{3} - 1}{ {x}^{2} + 2} \\ y' = \frac{(2 {x}^{3} - 1) '( {x}^{2} + 2) - ( {x}^{2} + 2)'(2 {x}^{3} - 1) }{( {x}^{2} + 2) {}^{2} } = \\ \frac{6 {x}^{2}( {x}^{2} + 2) - 2x(2x {}^{3} - 1)}{( {x}^{2} + 2) {}^{2} } = \\ \frac{6 {x}^{4} + 12 {x}^{2} - 4 {x}^{4} + 2x }{( {x}^{2} + 2) {}^{2} } = \\ \frac{2 {x}^{4} + 12x {}^{2} + 2x }{(x {}^{2} + 2) {}^{2} }$

8)

$y = \sqrt{ \sin(x) } \\ y' = \frac{1}{2 \sqrt{ \sin(x) } } \times \cos(x) = \frac{ \cos(x) }{2 \sqrt{ \sin(x) } }$