Question

Помогите пожалуйста, нужно найти производные ​

Answer 1

Ответ:

а)

$y' = 2x \sin(3x) + 3 \cos(3x) \times {x}^{2} = \\ = 2x \sin(3x) + 3 {x}^{2} \cos(3x)$

б)

формула:

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

$y't = 1 + \frac{1}{1 + 4 {t}^{2} } \times 2 \\ x't = 3 {t}^{2} + \frac{6}{1 + {t}^{2} }$

$y'x = \frac{1 + \frac{2}{1 + 4 {t}^{2} } }{3 {t}^{2} + \frac{6}{1 + {t}^{2} } } = \\ = \frac{1 + 4 {t}^{2} + 2}{1 + 4 {t}^{2} } \times \frac{1 + {t}^{2} }{3 {t}^{2} (1 + {t}^{2}) + 6 } = \\ = \frac{(1 + {t}^{2})(4 {t}^{2} + 3) }{(1 + 4 {t}^{2})(3 {t}^{4} + 3 {t}^{2} + 6) }$

при t = 1

$y'x = \frac{2 \times 7}{5 \times 12} = \frac{7}{30}$

в)

$y = {(tg( {x}^{3} ))}^{ ln(4x) }$

по формуле:

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

$( ln(y))' = ( ln(tg( { {x}^{3}) }^{ ln(4x) } )' = ( ln(4x) \times ln(tg( {x}^{3} ) )' = \\ = \frac{1}{4 x} \times 4 ln(tg( {x}^{3} ) ) + \frac{1}{tg( {x}^{3}) } \times \frac{1}{ { \cos}^{2}( {x}^{3}) } \times 3 {x}^{2} ln(4x) = \\ = \frac{ ln(tg( {x}^{3}) ) }{x} + \frac{ \cos( {x}^{3} ) }{ \sin( {x}^{3} ) } \times \frac{1}{ { \cos }^{2}( {x}^{3}) } \times 3 {x}^{2} ln(4x) = \\ = \frac{ ln(tg( {x}^{3} )) }{x} + \frac{3 {x}^{2} ln(4x) }{ \sin( {x}^{3} ) \cos( {x}^{3} ) }$

$y '= {(tg {x}^{3}) }^{ ln(4x) } \times ( \frac{ ln(tg( {x}^{3}) ) }{x} + \frac{3 {x}^{2} ln(4x) }{ \sin( {x}^{3} ) \cos( {x}^{3} ) }$