Question

Помогите!!!! даю 40 баллов!​

Answer 1

Ответ:

$f(x) = \frac{x}{ \sqrt{ {x}^{2} + 1} }$

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

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

$f'(0) = 0 - 0 = 0$