Question

Производные функции. Помните решить.​

Answer 1

Объяснение:

1.

$y '= 2 \times 5 {x}^{4} + 7 {x}^{6} - 0 = 10 {x}^{4} + 7 {x}^{6}$

$y' (- 1) = 10 + 7 = 17$

2.

$y' = ( {x}^{ - 4} + 3x - 2) '= - 4 {x}^{ - 5} + 3 = \\ = - \frac{4}{ {x}^{5} } + 3$

$y'(1) = - 4 + 3 = - 1$

3.

$y' = (4x - {x}^{ \frac{1}{2} } - 6) '= 4 - \frac{1}{2} {x}^{ - \frac{1}{2} } - 0 = \\ = 4 - \frac{1}{2 \sqrt{x} }$

$y'(1) = 4 - \frac{1}{2} = 4 - 0.5 = 3.5$

4.

$y' = ( {3}^{x} ) '\times ln(x) + (ln(x) )' \times {3}^{x} = \\ = ln(3) \times {3}^{x} ln(x) + \frac{1}{x} \times {3}^{x} = \\ = {3}^{x} ( ln(3) \times ln(x) + \frac{1}{x} )$

$y'(1) = 3( ln(3) \times ln(1) + 1) = \\ = 3(0 + 1) = 3$

5.

$y' = \frac{(5x - 2)'(3x - 1) - (3x - 1)'(5x - 2)}{ {(3x - 1)}^{2} } = \\ = \frac{5(3x - 1) - 3(5x - 2)}{ {(3x - 1)}^{2} } = \frac{15x - 5 - 15x + 6}{ {(3x - 1)}^{2} } = \\ = \frac{1}{ {(3x - 1)}^{2} }$

$y'(0) = \frac{1}{1} = 1$

6.

$y' = ( {e}^{x} + 3) '\times {x}^{2} + ( {x}^{2}) '\times ( e {}^{x} + 3) = \\ = {e}^{x} \times {x}^{2} + 2x(e {}^{x} + 3)$

$y'(0) = 0 + 0 = 0$

7.

$y' = \frac{4 {x}^{3}(1 - 3x) - ( - 3) \times {x}^{4} }{ {(1 - 3x)} {}^{2} } = \\ = \frac{4 {x}^{3} - 12 {x}^{4} + 3 {x}^{4} }{ {(1 - 3x)}^{2} } = \frac{4 {x}^{3} - 9 {x}^{4} }{ {(1 - 3x)}^{2} }$

$y '(- 2) = \frac{ 4 \times ( - 8) - 9 \times 16}{(1 + 6) {}^{2} } = \frac{ - 32 - 144}{49} = - \frac{176}{49}$