Question

Найдите производную функции, 20 баллов

Answer 1

Объяснение:

$1) \quad \qquad\small\: \: f(x) = 3 {x}^{6} + \large{ \tfrac{ 1}{4} } \small {x}^{4} - 2 {x}^{2} + 5x \\ \small f'(x)=3{\cdot}6x^{6-1} + \large{ \tfrac{ 1}{4} } \small {\cdot}4{x}^{4 - 1} - 2 {\cdot}2 {x}^{2 - 1} + 5 {x}^{1 - 1} = \\ = 18 {x}^{5} + {x}^{3} - 4x + 5$

$2) \: \: \: f(x) = (2 - 5) \cdot \sqrt{x} = - 3 \cdot {x}^{ \frac{1}{2} } \\ f'(x)=-3\cdot\frac{1}{2} \cdot{x^{\frac{1}{2}-1}} = - \frac{3}{2} \cdot{x^{ - \frac{1}{2}}} \\ f'(x) = - \frac{3}{2 \sqrt{x} }$

$3) \qquad\qquad \: f(x) = \frac{ {x}^{2} - 8x }{x + 2} \\ \small \: f'(x)= \frac{( {x}^{2} - 8x)' {\cdot}(x + 2) -( {x}^{2} - 8x){\cdot}(x + 2)' }{{(x + 2)}^{2} } = \\ \small = \frac{( 2{x} - 8) {\cdot}(x + 2) -( {x}^{2} - 8x ){\cdot}(1) }{{(x + 2)}^{2} } = \\ \small = \frac{( 2{x}^{2} - 8x + 4x - 16) -( {x}^{2} - 8x ) }{{(x + 2)}^{2} } = \\ \small = \frac{ 2{x}^{2} - 8x + 4x - 16 -{x}^{2} + 8x }{{(x + 2)}^{2} } = \\ \small = \frac{{x}^{2} + 4x - 16 }{{(x + 2)}^{2} } = \small \frac{({x}^{2} + 4x + 4 )- 20 }{{(x + 2)}^{2} } = \\ \small = \frac{({x} + 2)^{2} - 20 }{{(x + 2)}^{2} } = 1 - \frac{20}{ {(x + 2)}^{2} }$