Question

найти производную, даю 100 баллов, подробно ​

Answer 1

а)

$y = {x}^{5} tg(4x)$

$y '= ( {x}^{5} )'tg(4x) + (tg(4x))' \times {x}^{5} = \\ = 5 {x}^{4} tg(4x) + \frac{1}{ { \cos }^{2}(4x) } \times 4 \times {x}^{5}$

б)

$y'x = \frac{y't}{x't}$

$y't = 5 {t}^{4} + \frac{1}{1 + 16 {t}^{2} } \times 4$

$x't = \frac{5}{16} + \frac{1}{1 + 16 {t}^{2} }$

$y'x = \frac{5 {t}^{4} + \frac{4}{1 + 16 {t}^{2} } }{ \frac{5}{16} + \frac{1}{1 + 16 {t}^{2} } } = \\ = \frac{5 {t}^{4}(1 + 16 {t}^{2}) + 4 }{1 + 16 {t}^{2} } \times \frac{1 + 16 {t}^{2} }{5(1 + 16 {t}^{2} )+ 1 } = \\ = \frac{4 {t}^{4} + 80 {t}^{6} + 4 }{5 + 30 {t}^{2} + 1} = \frac{80 {t}^{6} + 4 {t}^{4} + 4 }{30 {t}^{2} + 6 } = \\ = \frac{2(40 {t}^{6} + 2 {t}^{4} + 2) }{2(15 {t}^{2} + 3)} = \frac{40 {t}^{6} + 2 {t}^{4} + 2 }{15 {t}^{2} + 3 }$

t = 1/2

$y'x(\frac{1}{2}) = \frac{40 \times \frac{1}{ {2}^{6} } + 2 \times \frac{1}{ {2}^{4} } + 2 }{15 \times \frac{1}{4} + 2 } = \\ = \frac{ \frac{40}{64} + \frac{1}{8} + 2 }{ \frac{15}{4} + 2} = \frac{ \frac{5}{8} + \frac{17}{8} }{ \frac{23}{4} } = \\ = \frac{22}{8} \times \frac{ 4 }{23} = \frac{11}{23}$

в)

$y = { (\cos( {x}^{4} )) }^{ \sin(8x) }$

по формуле:

$y' = ( ln(y))' \times y$

$( ln(y))'= ( ln( { \cos( {x}^{4} )) }^{ \sin(8x) } ) ' = ( \sin(8x) \times ln( \cos( {x}^{4} ) ) )' = \\ = 8 \cos(8x) \times ln( \cos( {x}^{4} ) ) + \frac{1}{ \cos( {x}^{4} ) } \times ( - \sin( {x}^{4} ) ) \times 4 {x}^{3} \sin(8x) = \\ = 8 \cos(8x) \times ln( \cos( {x}^{4} ) ) - 4 {x}^{3} tg( {x}^{4} ) \sin(8x)$

$y' = { \cos( {x}^{4} ) }^{ \sin(8x) } \times (8 \cos(8x) \times ln( \cos( {x}^{4} ) ) - 4 {x}^{3} tg( {x}^{4} ) \sin(8x) )$