Question

с дифференцированием функции

Answer 1

2.

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

$f'( \sqrt{2} ) = - \frac{1}{ \sqrt{2} \sqrt{2 - 1} } = - \frac{1}{ \sqrt{2} } = - \frac{ \sqrt{2} }{2} \\$

3

$f'(x) = \frac{(9x) '\sqrt{ {x}^{2} + 1} - ( {( {x}^{2} + 1)}^{ \frac{1}{2} } )'\times ( {x}^{2} + 1)' \times 9x }{ {x}^{2} + 1} = \\ = \frac{9 \sqrt{ {x}^{2} + 1 } - \frac{1}{2 \sqrt{ {x}^{2} + 1 } } \times 2x \times 9x}{ {x}^{2} + 1} = \\ = \frac{9 \sqrt{ {x}^{2} + 1} - \frac{9 {x}^{2} }{ \sqrt{ {x}^{2} + 1} } }{ {x}^{2} + 1 }$

$f'(2 \sqrt{2} ) = \frac{9 \sqrt{8 + 1} - \frac{9 \times 4 \times 2}{3} }{9} = \frac{27 - 24}{9} = \\ = \frac{3}{9} = \frac{1}{3}$