Question

Привет, помогите с алгеброй, пожалуйста! 50 баллов

Найдите производную функции:
1) f(x)=2x^5-x^3/3+3x^2-4

2) f(x)=(3x-5)√x

3) f(x)=x^2+9x/x-4

4) f(x)=2/x^3-3/x^6

Answer 1

Ответ:

1

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

$f'(x) = 2 \times 5 {x}^{4} - \frac{1}{3} \times 3 {x}^{2} + 3 \times 2x - 0 = \\ = 10 {x}^{4} - {x}^{2} + 6x$

2

$f(x) = (3x - 5) \sqrt{x}$

$f'(x) = (3x - 5)' \times \sqrt{x} + ( \sqrt{x} )' \times (3x - 5) = \\ = \sqrt{x} + \frac{1}{2} {x}^{ - \frac{1}{2} } \times (3x - 5) = \\ = \sqrt{x} + \frac{1}{2 \sqrt{x} } (3x - 5) = \sqrt{x} + 1.5 \sqrt{x} - \frac{5}{2 \sqrt{x} } = \\ = 2.5\sqrt{x} - \frac{5}{ 2\sqrt{x} }$

3

$f(x) = \frac{ {x}^{2} + 9x }{x - 4}$

$f'(x) = \frac{( {x}^{2} + 9x) '\times (x - 4) - (x - 4)' \times ( {x}^{2} + 9x)}{ {(x - 4)}^{2} } = \\ = \frac{(2x + 9)(x - 4) - ( {x}^{2} + 9x) }{ {(x - 4)}^{2} } = \frac{2 {x}^{2} - 8x + 9x - 36 - {x}^{2} - 9x }{ {(x - 4)}^{2} } = \\ = \frac{ {x}^{2} - 8x - 36}{ {(x - 4)}^{2} }$

4

$f(x) = \frac{2}{ {x}^{3} } - \frac{3}{ {x}^{6} } = 2 {x}^{ - 3} - 3 {x}^{ - 6 }$

$f'(x) = 2 \times ( - 3) {x}^{ - 4} - 3 \times ( - 6) {x}^{ - 7} = \\ = - \frac{6}{ {x}^{4} } + \frac{18}{ {x}^{7} }$