Question

Продиференціювати функцію:

Answer 1

Ответ:

17.8

$y = \sqrt{1 - 4 {x}^{2} } \times arcsin2x$

$y' = ( \sqrt{1 - 4 {x}^{2} } )' \times arcsin2x + (arcsin2x) '\times \sqrt{1 - 4 {x}^{2} } = \\ = \frac{1}{2} {(1 - 4 {x}^{2}) }^{ - \frac{1}{2} } \times (1 - 4 {x}^{2} )' \times arcsin2x + \frac{1}{ \sqrt{1 - 4 {x}^{2} } } \times 2 \times \sqrt{1 - 4 {x}^{2} } = \\ = \frac{1}{2 \sqrt{1 - 4 {x}^{2} } } \times ( - 8x) \times arcsin2x + 2 = \\ = - \frac{4arcsin2x}{ \sqrt{1 - 4 {x}^{2} } } + 2$

31.8

$y = \frac{4x}{4 + {x}^{2} }$

$y' = \frac{(4x)' \times (4 + {x}^{2} ) - (4 + {x}^{2} ) '\times 4x}{ {(4 + {x}^{2}) }^{2} } = \\ = \frac{4(4 + {x}^{2}) - 2x \times 4x }{ {(4 + {x}^{2} )}^{2} } = \\ = \frac{16 + 4 {x}^{2} - 8 {x}^{2} }{ {(4 + {x}^{2}) }^{2} } = \frac{16 - 4 {x}^{2} }{ {(4 + {x}^{2}) }^{2} }$

19.8

$y = \frac{ {2}^{arccos2x} }{2 {x}^{3} + 4x}$

$y' = \frac{( {2}^{arccos2x}) '\times (2 {x}^{3} + 4x) - (2 {x}^{3} + 4x) '\times {2}^{arccos2x} } { {(2 {x}^{3} + 4x)}^{2} } = \\ = \frac{ ln(2) \times {2}^{arccos2x} \times ( - \frac{1}{ \sqrt{1 - 4 {x}^{2} } }) \times 2(2 {x}^{3} + 4x) - (6 {x}^{2} + 4) \times {2}^{arccos2x} }{ {(2 {x}^{3} + 4x) }^{2} } = \\ = \frac{ {2}^{arccos2x} ( - \frac{2 (2 {x}^{3} + 4x) ln(2) }{ \sqrt{1 - 4 {x}^{2} } } - (6 {x}^{2} + 4)) }{ {(2 {x}^{3} + 4x) }^{2} } = \\ = - \frac{ {2}^{arccos2x} }{ {(2 {x}^{3} + 4x) }^{2} } \times ( \frac{2(2 {x}^{3} + 4x) ln(2) }{ \sqrt{1 - 4 {x}^{2} } } + 6 {x}^{2} + 4)$