Question

Найдите
производные следующих функций(пожалуйста с фото!!!)
1) y= x^2*(1+7sinx)
2. y= x*arctgx
3. y= 4-3x^2+x^4/x^3
4. y=8arcsinx+5arccosx-11

ДАЮ 25 БАЛЛОВ!

Answer 1

Ответ:

1

$y' = ( {x}^{2} )'(1 + 7 \sin(x) ) + (1 + 7 \sin(x) )' \times {x}^{2} = \\ = 2x(1 + 7 \sin(x)) + 7 \cos(x) \times {x}^{2} = \\ = 14x \sin(x) + 2x + 7 {x}^{2} \cos(x)$

2

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

3

$y' = \frac{(4 - 3 {x}^{2} + {x}^{4})'( {x}^{3} ) - ( {x}^{3} )'(4 - 3 {x}^{2} + {x}^{4} ) }{ {( {x}^{3} )}^{2} } = \\ = \frac{( - 6x + 4 {x}^{3}) \times {x}^{3} - 3 {x}^{2}(4 - 3 {x}^{2} + {x}^{4}) }{ {x}^{6} } = \\ = \frac{ - 6 {x}^{4} + 4 {x}^{6} - 12 {x}^{2} + 9 {x}^{4} - 3 {x}^{6} }{ {x}^{6} } = \\ = \frac{ {x}^{6} + 3 {x}^{4} - 12 {x}^{2} }{ {x}^{6} } = \frac{ {x}^{4} + 3 {x}^{2} - 12 }{ {x}^{4} }$

4

$y' = \frac{8}{ \sqrt{1 - {x}^{2} } } - \frac{5}{ \sqrt{1 - {x}^{2} } } - 0 = \\ = \frac{3}{ \sqrt{1 - {x}^{2} } }$