Question

Решите с объяснением

Answer 1

Ответ:

$1) {({4}^{x})}^{l} = {4}^{x} ln(4) \\ {({x}^{- 2} )}^{l} = - 2 \times {x}^{ - 2 - 1} = - 2 {x}^{ - 3} \\ y = {4}^{x} ln(4) - 2 {x}^{ - 3}$

$2){( {3}^{x} )}^{l} = {3}^{x} ln(3) \\ {( {2x}^{ - 2}) }^{l} = 2 \times ( -2) \times {x}^{ - 2 - 1} = - 4 {x}^ {- 3} \\ {(3 log_{2}(x) )}^{l} = \frac{3}{x ln(2) } \\ {f}^{l} (x) = {3}^{x} ln(3) - \frac{4}{ {x}^{3} } + \frac{3}{x ln(2) }$

$3)( {x}^{2} - 1)( {x}^{2} + 1) = {x}^{4} - 1\\ {f}^{l} (x) = 4 {x}^{4 - 1} = 4 {x}^{3}$

$4){f}^{l} (x) = \frac{2 \times 2 {x}^{2 - 1} ( {x}^{3} + 3 ) - 2 {x}^{2} \times 3 \times {x}^{3 - 1} }{ {( {x}^{3} + 3 )}^{2} } \\ \frac{4x( {x}^{3} + 3) - 6 {x}^{4} }{ {( {x}^{3} + 3)}^{2} } = \\ = \frac{12x - 2 {x}^{4} }{ {( {x}^{3} + 3 )}^{2} }$

$5) {f}^{l} (x) ={ ((5x + 2)2cosx) }^{l} = \\ = { ((10x cosx + 4cosx) }^{l} \\ \\ {(10xcosx) }^{l} = 10cosx + 10x \times ( - sinx) = \\ = 10cosx - 10xsinx \\ \\ {(4 cosx)}^{l} = - 4sinx \\ \\ {f}^{l} (x) =10cosx - 10xsinx - 4sinx$