Question

Необходимо решить, заменив N=10

Answer 1

Ответ:

$1)y = {( {x}^{10} - 20x + 10)}^{11}$

$y' = 11 {( {x}^{10} - 20x + 10)}^{10} \times (10 {x}^{9} - 20)$

$4)y = \sqrt[10]{tg(x)} \\ y' = \frac{1}{10} {(tg(x))}^{ - \frac{9}{10} } \times \frac{1}{ { \cos( x ) }^{2} } = \frac{1}{10 { \cos(x) }^{2} \sqrt[10]{ {tg(x)}^{9} } }$

$2)y = { \cos(x) }^{10}$

$y' = 10 { \cos(x) }^{9} \times ( - \sin(x) )$

$3)y = \sin( \sqrt[10]{x} )$

$y' = \cos( \sqrt[10]{x} ) \times \frac{1}{10 \ \sqrt[10]{ {x}^{9} } }$

$5)y = ln( \sqrt[11]{x} )$

$y' = \frac{1}{ \sqrt[11]{x} } \times \frac{1}{11 \sqrt[11]{ {x}^{10} } } = \frac{1}{11x}$

$6)y = ctg( {x}^{10} ) \\ y' = - \frac{1}{ { \sin( {x}^{10} ) }^{2} } \times 10 {x}^{9}$

$7)y = arcsin(10x) \\ y' = \frac{10}{ \sqrt{1 - 100 {x}^{2} } }$

$8)y = \sqrt[10]{arctg(x)}$

$y' = \frac{1}{10 \sqrt[10]{ {arctg(x)}^{9} } } \times \frac{1}{1 + {x}^{2} }$

$9)y = {e}^{10x} + {11}^{10x} \\ y' = 10 {e}^{10x} + 10 ln(11) \times {11}^{10x} \times$