Question

вычислить значение производной срочно ​

Answer 1

Ответ:

1.

$f'(x) = 2x \\ f'(3) = 2 \times 3 = 6$

2.

$f'(x) = 2 \times 3 {x}^{2} = 6 {x}^{2} \\ f'(-2) = 6 \times {( - 2)}^{2} = 6 \times 4 = 24$

3.

$f'(x) = \frac{1}{2} {x}^{ - \frac{1}{2} } = \frac{1}{2 \sqrt{x } } \\ f'(4) = \frac{1}{2 \times \sqrt{4} } = \frac{1}{4} = 0.25$

4.

$f'(x) = 2x - 3 \\ f'( \frac{1}{2} ) = 1 - 3 = - 2$

5.

$f'(x) = ( {x}^{3} (2x + {x}^{2} )) '= (2 {x}^{4} + {x}^{5} )' = \\ = 8 {x}^{3} + 5 {x}^{4} \\ f'(1) = 8 + 5 = 13$

6.

$f'(x) = ( {x}^{ - 1} ) '= - {x}^{ - 2} = - \frac{1}{ {x}^{2} } \\ f'(5) = - \frac{1}{25}$

7.

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

8.

$f'(x) = 7 \times \frac{3}{5} {x}^{ - \frac{2}{5} } = \frac{21}{5 \sqrt[5]{ {x}^{2} } }$

производной в точке 0 не существует, т к знаменатель не равен 0.

9.

$f'(x) = \frac{(3 - x)'(2 + x) - (2 + x)'(3 - x)}{ {(2 + x)}^{2} } = \\ = \frac{ - (2 + x) - 1 \times (3 - x)}{ {(2 + x)}^{2} } = \frac{ - 2 - x - 3 + x}{ {(2 + x)}^{2} } = \\ = - \frac{5}{ {(x + 2)}^{2} } \\ f'(3) = - \frac{5}{ {5}^{2} } = - \frac{1}{5} = - 0.2$

10.

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