Question

с решением самостоятельной работы.

Answer 1

1.

а

$f'(x) = \frac{1}{4} \times 4 {x}^{3} - 2 \times \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = {x}^{3} - \frac{1}{ \sqrt{x} }$

б

$f'(x) = ( {x}^{2} )' \times \cos(x) + ( \cos(x) )' \times {x}^{2} = \\ = 2x \cos(x) - \sin(x) \times {x}^{2}$

в

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

2.

$y' = 4 \sin {}^{3} (x) \times ( \sin(x)) ' + 4 \cos {}^{3} (x) \times ( \cos(x)) ' = \\ = 4 \sin {}^{3} (x) \times \cos(x) + 4\cos {}^{3} (x) \times ( - \sin(x)) = \\ = 4 \sin(x) \cos(x) \times ( \sin {}^{2} (x) - \cos {}^{2} (x) ) = \\ = 2 \sin(2x) \times ( - \cos(2x)) = - \sin(4x) \\ \\ y'( \frac{\pi}{4} ) = - \sin(4 \times \frac{\pi}{4} ) = - \sin(\pi) = 0$