Question

найти производную: (10 класс)
F(x)=(2x^3-1)*(x^2-1)
F(1)
=================
F(x)=дробь X^2-1/x^2-1
F(-2)
=================
f(x)=e^x*(x^2-5x+1)
f(0)

Желательно с обьяснением!
Спасибо!

Answer 1

Ответ:

1.

$f(x) = (2 {x}^{3} - 1)( {x}^{2} - 1)$

используем формулу:

$(uv)' = u'v + v'u$

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

$f'(1) = 10 - 6 - 2 = 2$

2.

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

$f'(x) =$

$f'( - 2) =$

3.

$f(x) = {e}^{x} ( {x}^{2} - 5x + 1)$

используем формулу из 1)

$f'(x) = ( {e}^{x} )'( {x}^{2} - 5x + 1) + ( {x}^{2} - 5x + 1) ' \times {e}^{x} = \\ = {e}^{x} ( {x}^{2} - 5x + 1) + (2x - 5) \times {e}^{x} = \\ = {e}^{x} ( {x}^{2} - 5x + 1 + 2x - 5) = \\ = {e}^{x} ( {x}^{2} - 3x - 4)$

$f'(0) = {e}^{0} (0 - 4) = 1 \times ( - 4) = - 4$