Question

СРОЧНО!!!!
Знайдіть похідні
слідуючих
функцій:
1) у=-5,23
2)у=х⁴
3)у=3^х
4)у=sinx+3x²
5)y=e^x-3*in x+7²
6)y=1:x²+4inx+⁷✓x³
7)y=x⁵*tgx
8)y=5x+2:3x²-1
9)y=inx:x⁵
10)y=in(x⁴-x)

^–степінь​

Answer 1

Ответ:

1.

$y= - 5.23$

$y' = 0$

2.

$y = {x}^{4} \\ y' = 4 {x}^{3}$

3.

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

4.

$y = \sin(x) + 3 {x}^{2} \\ y' = \cos(x) + 6x$

5.

$y = {e}^{x} - 3 ln(x) + {7}^{2}$

$y '= {e}^{x} - \frac{3}{x}$

6.

$y = \frac{1}{ {x}^{2} } + 4 ln(x) + \sqrt[7]{ {x}^{3} } = \\ = {x}^{ - 2} + 4 ln(x) + {x}^{ \frac{3}{7} }$

$y' = - 2 {x}^{ - 3} + \frac{4}{x} + \frac{3}{7} {x}^{ - \frac{4}{7} } = \\ = - \frac{2 {}^{} }{ {x}^{3} } + \frac{4}{x} + \frac{3}{7 \sqrt[7]{ {x}^{4} } }$

7.

$y = {x}^{5} tgx$

$y' = ( {x}^{5} )'tgx + (tgx)' \times {x}^{5} = \\ = 5 {x}^{4} tgx + \frac{ {x}^{5} }{ \cos {}^{2} (x) }$

8.

$y = \frac{5x + 2}{3 {x}^{2} - 1}$

$y' = \frac{(5x + 2)'(3 {x}^{2} - 1) - (3 {x}^{2} - 1)'(5x + 2) }{ {(3 {x}^{2} - 1)}^{2} } = \\ = \frac{5(3 {x}^{2} - 1) - 6x(5x + 2)}{ {(3 {x}^{2} - 1) }^{2} } = \\ = \frac{15 {x}^{2} - 5 - 30 {x}^{2} - 12x}{ {(3 {x}^{2} - 1) }^{2} } = \frac{ - 30 {x}^{2} - 12x - 5}{ {(3 {x}^{2} - 1)}^{2} } = \\ = - \frac{ 30 {x}^{2} + 12x + 5}{ {(3 {x}^{2} - 1) }^{2} }$

9.

$y = \frac{ ln(x) }{ {x}^{5} }$

$y '= \frac{( ln(x)) ' \times {x}^{5} - ( {x}^{5} )' \times ln(x) }{ {x}^{10} } = \\ = \frac{ \frac{1}{x} \times {x}^{5} - 5 {x}^{4} ln(x) }{ {x}^{10} } = \frac{ {x}^{4} - 5 {x}^{4} ln(x) }{ {x}^{10} } = \\ = \frac{1 - 5 ln(x) }{ {x}^{6} }$

10

$y = ln( {x}^{4} - x )$

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