Question

Помогите пожалуйста решить срочно.

Answer 1

Ответ:

1)

${((3x + 4)}^{2} )' = {7(3x + 4)}^{6} \times (3x + 4)' = 7 {(3x + 4)}^{6} \times 3 = 21 {(3x + 4)}^{6}$

2)

$( {e}^{2 {x}^{4} - 5 \sin(x) } )' = (2{x}^{4} - 5 \sin(x) )' \times {e}^{2 {x}^{4} - 5 \sin(x) } = (8 {x}^{3} - 5 \cos(x) ) \times {e}^{2 {x}^{4} - 5 \sin(x) } = 8 {x}^{3} \times {e}^{2 {x}^{4} - 5 \sin(x) } - 5 \cos(x) \times {e}^{2 {x}^{4} - 5 \sin(x) }$

3)

$( \cos(1 + 6 {x}^{7} - 13x ) )' = - \sin(1 + 6 {x}^{7} - 13x) \times (1 + 6 {x}^{7} - 13x)' = - \sin(1 + 6 {x}^{7} - 13x ) \times (42 {x}^{6} - 13)$

4)

$log_{4}(2 ln(x) - 7 \tan(x) + 9 ) = log_{ {2}^{2} }(2 ln(x) - 7 \tan(x) + 9 ) = (\frac{1}{2} \times log_{2}(2 ln(x) - 7 \tan(x) + 9 ) )'$

$\frac{1}{2} \times \frac{1}{ ln(2)(2 ln(x) - 7 \tan(x) + 9) } \times (2 ln(x) - 7 \tan(x) + 9)' = \frac{1}{2} \times \frac{1}{ ln(2)(2 ln(x) - 7 \tan(x) + 9) } \times ( \frac{2}{x} - \frac{7}{ \cos ^{2} (x) } )$

5)

$( ln(9 {x}^{3} + 8x - 5 \sqrt{x} ))' = \frac{1}{9 {x}^{3} + 8x - 5 \sqrt{x} } \times (9 {x}^{3} + 8x - 5 \sqrt{x} )' = \frac{1}{9 {x}^{3} + 8x - 5 \sqrt{x} } \times (27 {x}^{2} + 8 - \frac{5}{2 \sqrt{x} } )$