Найдите производную функции 1) y=-8x+x^12
2) y=9x^2+5x^4+15
3) y=9x-3√x
4) y=1/x-7x^-4+10
5) y=-2/x^4-3sinx
6)y=tgx+√x
7)y=(x^4+7)(1+x^5)
8)y=5/x^6+ctgx
9)y=√x(3-4x)
10) y=x^8cosx
11) y=sinx/4x^3
заранее

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Ответ:

Антонио789176

06.03.2021 09:51

$y' = - 8 + 12 {x}^{11}$

$y '= 18x + 20 {x}^{3} + 0 = 18x + 20 {x}^{3}$

$y' = (9x - 3 {x}^{ \frac{1}{2} } )' = 9 - 3 \times \frac{1}{2} {x}^{ - \frac{1}{2} } ) = \\ = 9 - \frac{3}{2 \sqrt{x} }$

$y' = ( {x}^{ - 1} - 7 {x}^{ - 4} + 10)' = \\ = - {x}^{ - 2} + 28 {x}^{ - 5} + 0= \\ = - \frac{1}{ {x}^{2} } + \frac{28}{ {x}^{5} }$

$y' = ( - 2 {x}^{ - 4} - 3 \sin(x)) '= \\ = 8 {x}^{ - 5} - 3 \cos(x) = \\ = \frac{8}{ {x}^{5} } - 3 \cos(x)$

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

$y' = ( {x}^{4} + 7)'(1 + {x}^{5} ) + (1 + {x}^{5} )'( {x}^{4} + 7) = \\ = 4 {x}^{3} ( {x}^{5} + 1) + 5 {x}^{4} ( {x}^{4} + 7) = \\ = 4 {x}^{8} + 4 {x}^{3} + 5 {x}^{8} + 35 {x}^{4} = \\ = 9 {x}^{8} + 35 {x}^{4} + 4 {x}^{3}$

$y' = (5 {x}^{ - 6} + ctgx)' = - 30 {x}^{ - 7} - \frac{1}{ { \sin}^{2} x} = \\ = - \frac{30}{ {x}^{7} } - \frac{1}{ { \sin }^{2}x }$

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

10.

$y' = ( {x}^{8} )' \cos(x) + ( \cos(x))' \times {x}^{8} = \\ = 8 {x}^{7} \cos(x) - \sin(x) \times {x}^{8}$

11.

$y '= \frac{( \sin(x)) ' \times 4 {x}^{3} - (4 {x}^{3} )' \times \sin(x) }{16 {x}^{6} } = \\ = \frac{ \cos(x) \times 4 {x}^{3} - 12 {x}^{2} \times \sin(x) }{16 {x}^{6} } = \\ = \frac{4 {x}^{2}(x \cos(x) - 3\sin(x)) }{16 {x}^{6} } = \\ = \frac{x \cos(x) - 3 \sin(x) }{4 {x}^{4} }$