Question

Помогите пожалуйста, нужно найти производную ​

Answer 1

Ответ:

а)

$y' = 7 {x}^{6} ctg(10x) - \frac{1}{ { \sin}^{2}(10x) } \times 10 \times {x}^{7} = \\ = 7 {x}^{6} ctg(10x) - \frac{10 {x}^{7} }{ { \sin}^{2}(10x) }$

б)

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

$y't = 2t - \frac{1}{ \sqrt{1 - 25 {t}^{2} } } \times 5$

$x't = \frac{2}{25} - \frac{1}{ \sqrt{1 - 25 {t}^{2} } } \times 5$

$y'x = \frac{2t - \frac{5}{ \sqrt{1 - 25 {t}^{2} } } }{ \frac{2}{25} - \frac{5}{ \sqrt{1 - 25 {t}^{2} } } } = \\ = \frac{2t \sqrt{1 - 25 {t}^{2} } - 5}{ \sqrt{1 - 25 {t}^{2} } } \times \frac{ \sqrt{1 - 25 {t}^{2} } }{2 \sqrt{1 - 25 {t}^{2} } - 5} = \\ = \frac{2t \sqrt{1 - 25 {t}^{2} } - 5}{2 \sqrt{1 - 25 {t}^{2} } - 5 }$

при t = 1/25

$y'x( \frac{1}{25} ) = \frac{ \frac{2}{25} \sqrt{1 - \frac{25}{625} } - 5}{2 \sqrt{1 - \frac{25}{625} } - 5 } = \\ = \frac{ \frac{2}{25} \times \sqrt{ \frac{25 - 1}{25} } - 5}{2 \times \sqrt{ \frac{24}{25} } - 5} = \\ = \frac{ \frac{2}{25} \times \frac{ \sqrt{24} }{5} - 5}{2 \times \frac{ \sqrt{24} }{5} - 5 } = \\ \frac{2 \sqrt{24} - 625 }{125} \times \frac{5}{2 \sqrt{24} - 25} = \\ = \frac{2 \sqrt{24} - 625 }{25(2 \sqrt{24} - 25) }$

в)

$y = {( \sin( {x}^{2} )) }^{ \sin(7x) }$

формула:

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

$( ln(y))' = ( ln( { \sin( {x}^{2} ) }^{ \sin(7x) } ) )' = ( \sin(7x) \times ln( \sin( {x}^{2} ) ) ) '= \\ = 7 \cos(7x) \times ln( \sin( {x}^{2} ) ) + \frac{1}{ \sin( {x}^{2} ) } \times \cos( {x}^{2} ) \times 2x \times \sin(7x) = \\ = 7 \cos(7x) ln( \sin( {x}^{2} ) ) + 2x \times ctg( {x}^{2} ) \sin(7x)$

$y' = {( \sin( {x}^{2} )) }^{ \sin(7x) } \times (7 \cos(7x) ln( \sin( {x}^{2} ) ) + 2x \times ctg( {x}^{2} ) \times \sin(7x))$