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vikabobkova2010

07.04.2020

$f'(x) = - 5 \times 2x + 0 - \frac{1}{2} \times ( {x}^{ - 1} )' = \\ = - 10x + \frac{1}{2} {x}^{ - 2} = - 10x + \frac{1}{2 {x}^{2} }$

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

$f'(x) = (( { \frac{1}{4x} - 2)}^{ \frac{1}{2} } )' = \frac{1}{2} {( \frac{1}{4x} - 2)}^{ - \frac{1}{2} } \times ( \frac{1}{4} {x}^{ - 1} - 2)' = \\ = \frac{1}{2 \sqrt{ \frac{1}{4x} - 2 } } \times ( - { \frac{1}{4} } {x}^{ - 2} ) = \frac{1}{2} \times \sqrt{ \frac{4x}{1 - 8x} } \times ( - \frac{1}{4 {x}^{2} } ) = \\ = - \frac{1}{8 {x}^{2} } \times \frac{2 \sqrt{x} }{ \sqrt{1 - 8x} } = - \frac{1}{4x \sqrt{x(1 - 8x)} }$