Question

Помогите решить все. Даю 50 баллов!!!

Answer 1

Ответ:

$1)y = { ( log_{4}(2x + 3)) }^{arcsinx}$

По формуле:

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

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

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

$2)y = \frac{ { \cos(7x - 1) }^{4} }{lg(x + 5)}$

$y' = \frac{4 { \cos(7x - 1) }^{3} \times ( - \sin(7x - 1) ) \times 7 \times lg(x + 5) - \frac{1}{ ln(10) \times (x + 5)} \times { \cos(7x - 1) }^{4} }{ {lg}^{2} (x + 5)} = \frac{ { \cos(7x - 1) }^{3} ( - 28 \sin(7x - 1) \times lg(x + 5) - \frac{ \cos(7x - 1) }{ ln(10) \times (x + 5) } )}{ {lg}^{2}(x + 5) }$

$3) y= \frac{ \sqrt{ {(x + 2)}^{3} } {(x - 1)}^{4} }{ {(x + 2)}^{2} }$

по той же формуле, что в 1)

$(ln(y))' = ( ln( \frac{ {(x + 2)}^{ \frac{3}{2} } {(x - 1)}^{4} }{ {(x + 2)}^{2} } ) )' = ( ln( {(x + 2)}^{ \frac{3}{2} } + ln( {(x - 1)}^{4} ) - ln( {(x + 2)}^{2} ) )' = \frac{3}{2} \times \frac{1}{x + 2} + \frac{4}{x - 1} - \frac{2}{x + 2} = \frac{3(x - 1)(x + 2) + 8(x + 2)(x + 2) - 4(x - 1)(x + 2)}{2(x + 2)(x - 1)(x + 2)} = \frac{7 {x}^{2} + 31x + 34 }{2(x - 1) {(x + 2)}^{2} }$

$y = \frac{ \sqrt{ {(x + 2)}^{3} } {(x - 1)}^{4} }{ {(x + 2)}^{2} } \times \frac{7 {x}^{2} + 31x + 34}{2 {(x + 2)}^{2}(x - 1) } = \frac{(7 {x}^{2} + 31x + 34) {(x - 1)}^{3} }{2 \sqrt{ {(x + 2)}^{5} } }$

$4)y = \frac{4 log_{ 3}(3x + 1) }{ {(x + 1)}^{2} }$

$y' = \frac{ \frac{4}{ ln(3) \times (3x + 1)} \times 3 {(x + 1)}^{2} - 2(x + 1) \times 4 log_{3}(3x + 1) }{ {(x + 1)}^{4} } = \frac{(x + 1)( \frac{12(x + 1)}{ ln(3) \times (3x + 1)} - 8 log_{3}(3x + 1)) }{ {(x + 1)}^{4} } = \frac{ \frac{12(x + 1)}{ ln(3) \times ( 3x + 1)} - 8 log_{3}(3x + 1) }{ {(x + 1)}^{3} }$