Question

решить 3,4 и 5 пример. Пытался сам, но не получалось. Тема : Производная сложной функции.

Answer 1

3.

$y = \sqrt{ \frac{2}{x} + 3 {x}^{2} + 25} = {( \frac{2}{x} + 3 {x}^{2} + 25) }^{ \frac{1}{2} } = \\ = {(2 {x}^{ - 1} + 3 {x}^{2} + 25)}^{ \frac{1}{2} }$

$y' = \frac{1}{2} {(2 {x}^{ - 1} + 3 {x}^{2} + 25x) }^{ - \frac{1}{2} } \times (2 {x}^{ - 1} + 3 {x}^{2} + 25x)' = \\ = \frac{1}{2 \sqrt{ \frac{2}{x} + 3 {x}^{2} + 25x} } \times ( - 2 {x}^{ - 2} + 6x + 25) = \\ = \frac{25 + 6x - \frac{2}{ {x}^{2} } }{2 \sqrt{ \frac{2}{x} + 3 {x}^{2} + 25x } }$

4.

$f(t) = \frac{1}{8 {t}^{3} - 12 t ^{2} + 9 } = {(8 {t}^{3} - 12 {t}^{2} + 9) }^{ - 1} \\$

$f'(t) = - {(8 {t}^{3} - 12 {t}^{2} + 9) }^{ - 2} \times (8 {t}^{3} - 12 {t}^{2} + 9)' = \\ = - \frac{16 {t}^{2} - 24t }{ {(8 {t}^{3} - 12 {t}^{2} + 9) }^{2} }$

5.

$y = {( \frac{3x - 5}{8 {x}^{2} + 9 } })^{2} \\$

$y '= 2 \times \frac{3x - 5}{8 {x}^{2} + 9 } \times ( \frac{3x - 5}{8 {x}^{2} + 9} )' = \\ = \frac{2(3x - 5)}{8 {x}^{2} + 9 } \times \frac{(3x - 5)'(8 {x}^{2} + 9) - (8 {x}^{2} + 9)'(3x - 5) }{ {(8 {x}^{2} + 9)}^{2} } = \\ = \frac{2(3x - 5)}{8 {x}^{2} + 9 } \times \frac{3(8 {x}^{2} + 9) - 16x(3x - 5) }{ {(8 {x}^{2} + 9)}^{2} } = \\ = \frac{2( 3x - 5)}{8 {x}^{2} + 9} \times \frac{24 {x}^{2} + 27 - 48 {x}^{2} + 80x }{ {(8 {x}^{2} + 9)}^{2} } = \\ = \frac{2(3x - 5)( - 24 {x}^{2} + 80x + 27) }{ {(8 {x}^{2} + 9) }^{3} }$