Question

Продифференцировать функции. Алгебра 10 класс. Решите, пожалуйста , с подробным решением. Буду признательно благодарен вам.

Answer 1

Ответ:

1.

$y' = 4 {x}^{3} - {x}^{2} + 5x - 0.3$

2.

$y '= (2x - 3)( {x}^{2} + 2x - 1) + (2x + 2)( {x}^{2} - 3x + 3) = \\ = 2 {x}^{3} + 4 {x}^{2} - 2x - 3 {x}^{2} - 6x + 3 + 2 {x}^{3} - 6 {x}^{2} + 6x + 2 {x}^{2} - 6x + 6 = \\ 4 {x}^{3} - 3 {x}^{2} - 8x + 9$

3.

$y' = \frac{1}{2 \sqrt{x} } ( \frac{1}{ \sqrt{x} } - 1) - \frac{1}{2x \sqrt{x} } ( \sqrt{x} + 1) = \\ = \frac{1}{2x} - \frac{1}{2 \sqrt{x} } - \frac{1}{2x} - \frac{1}{2x \sqrt{x} } = \\ = - \frac{1}{2 \sqrt{x} } - \frac{1}{2x \sqrt{x} }$

4.

$y' = arcsinx + \frac{x}{ \sqrt{1 - {x}^{2} } }$

5.

$y' = {e}^{x} + x {e}^{x} = {e}^{x} (x + 1)$

6.

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

7.

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

8.

$y' = \frac{ \cos(t)(1 + \cos(t)) + \sin(t) \times \sin(t) }{ {(1 + \cos(t)) }^{2} } = \\ = \frac{ \cos(t) + { \cos }^{2}( t) + { \sin }^{2} (t)}{ {(1 + \cos(t)) }^{2} } = \\ = \frac{1 + \cos(t) }{ {(1 + \cos(t)) }^{2} } = \frac{1}{1 + \cos(t) }$

9.

$y '= \frac{ \frac{1}{ \sqrt{1 - {x}^{2} } }arccosx + \frac{1}{ \sqrt{1 - {x}^{2} } } arcsinx }{ {arccos}^{2} x} = \\ = \frac{1}{ \sqrt{1 - {x}^{2} } } ( \frac{1}{arccosx} + \frac{arsinx}{ {arccos}^{2}x } )$

10.

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