Question

знайдіть похідну функції

Answer 1

1.1

1

$y '= 4 {x}^{3} - 6 {x}^{2} + 1$

2

$y' = ((x - 1)(2x + 1)) '= (2 {x}^{2} + x - 2x - 1) '= \\ = (2 {x}^{2} - x - 1)' = 4x - 1$

3

$y' = \frac{( {x}^{2} - 1) '\sin(x) - ( \sin(x)) ' \times ( {x}^{2} - 1) }{ \sin {}^{2} (x) } = \\ = \frac{2x \sin(x) - ( {x}^{2} - 1) \cos(x) }{ \sin {}^{2} (x) }$

4

$y' = \frac{(1 - 2x)'(2 {x}^{2} + 3) - (2 {x}^{2} + 3)'(1 - 2x) }{ {(2 {x}^{2} + 3)}^{2} } = \\ = \frac{ - 2(2 {x}^{2} + 3) - 4x(1 - 2x) }{ {(2 {x}^{2} + 3)}^{2} } = \\ = \frac{ - 4 {x}^{2} - 6 - 4x + 8 {x}^{2} }{ {(2 {x}^{2} + 3) }^{2} } = \frac{4 {x}^{2} - 4x - 6 }{(2 {x}^{2} + 3) {}^{2} }$

1.2

$f'(x) = {x}^{3} - 2x \\ f'(2) = 8 - 4 = 4$

2.1

$s(t) = {t}^{3} + 3t {}^{2}$

$v(t) = s'(t) = 3 {t}^{2} + 6 t \\ v(1) = 3 + 6 = 9$

2.2

$s(t) = {t}^{2} - 4t + 6 \\ v(t) = 2t - 4 \\ \\ 2t - 4 = 10 \\ 2t = 14\\ t = 7$

3.1

$y = {x}^{2} - 4x \\ x_0 = 2$

$f(x) = y(x_0) + y'(x_0)(x - x_0)$

$y(2) = 4 - 8 = - 4$

$y' = 2x - 4$

$y'(2) = 4 - 4 = 0$

$f(x) = - 4 + 0(x - 2) = - 4$