Question

30 баллов!!!!
помогите пожалуйста найти производные функций,используя таблицу производных и правила дифференцирования
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Answer 1

Ответ:

$1)y' = 24 {x}^{5} - 8 {x}^{2} + 3 \times \frac{1}{2} {x}^{ - \frac{1}{2} } = 24 {x}^{5} - 8 {x}^{2} + \frac{3}{2 \sqrt{x} }$

$2)y' = \frac{ - {x}^{2} - 2x(7 - x) }{ {x}^{4} } = \frac{ - {x}^{2} - 14x + 2 {x}^{2} }{ {x}^{4} } = \frac{ {x}^{2} - 14x}{ {x}^{4} } = \frac{x - 14}{ {x}^{3} }$

$3)y' = (2x - 3) \times {3}^{x} + ln(3) \times {3}^{x} ( {x}^{2} - 3x + 4) = {3}^{x} (2x - 3 + ( {x}^{2} + 3x + 4) ln(3) )$

$4)y' = \frac{1}{5} {x}^{ - \frac{4}{5} } ln(x) + \frac{1}{x} \times \sqrt[5]{x} = \frac{ ln(x) }{5 \sqrt[5]{ {x}^{4} } } + \frac{1}{ \sqrt[5]{ {x}^{4} } } = \frac{ ln(x) + 5 }{5 \sqrt[5]{ {x}^{4} } }$

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

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

$7)y' = \frac{ \frac{1}{2} {x}^{ - \frac{1}{2}}( \sqrt{x} + 1) - \frac{1}{2} {x}^{ - \frac{1}{2} } \sqrt{x} }{ {( \sqrt{x} + 1)}^{2} } = \frac{ \frac{1}{2 \sqrt{x} } ( \sqrt{x} + 1 - \sqrt{x} ) }{ {( \sqrt{x} + 1) }^{2} } = \frac{1}{2 \sqrt{x} {( \sqrt{x} + 1)}^{2} }$

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

$9)y' = \frac{1}{2} {(1 - {x}^{2} )}^{ - \frac{1}{2} } \times ( - 2x) + \cos( \frac{x}{2} ) \times \frac{1}{2} = - \frac{x}{ \sqrt{1 - {x}^{2} } } + \frac{1}{2} \cos( \frac{x}{2} )$

$10)y' = 3arcctg( \frac{x}{3} ) - \frac{3x}{1 + \frac{ {x}^{2} }{9} } - {(9 + {x}^{2} )}^{ - 2} \times 2x = 3arcctg( \frac{x}{3} ) - \frac{3x}{1 + \frac{ {x}^{2} }{9} } - \frac{2x}{ {(9 + {x}^{2} )}^{2} }$