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khokhlovakater
@khokhlovakater
July 2022
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СРОЧНО!АЛГЕБРА 10 КЛАСС! ДАЮ 80Б.ТОЛЬКО С РЕШЕНИЕМ
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konstantsya
1) f'(x) = (x^3)' = 3x^(3-1) = 3x^2
2) f'(x) = (3x)' = 3
3) f'(x) = (x^(-8))' = -8x^(-9)
4) f'(x) = (3,7)' = 0
5) f'(x) = (1/x)' = (x^(-1))' = -x^(-2) = -1/x^2
6) f'(x) = (3x^8-2x^6+4x^4-3x^2+x-8)' =
= 3*8x^(8-1)-2*6x^(6-1)+4*4x^(4-1)-3*2x^(2-1)+x^0-0 =
= 24x^7-12x^5+16x^3-6x+1
7) 2) f'(x) = (x^4+
)' = (x^4+x^(1/2) )' =
= 4x^(4-1)+1/2 *x^(1/2-1) = 4x^3-1/2*x^-1/2
4x^3-
8) f'(x) = (1/4 *x^(-4))' = 1/4*(-4)*x^(-4-1) = -x^(-5)
9) 6) f'(x) = (x^8-3x^4-x+5)' = 8x^(8-1)-3*4x^(4-1)-1X^0+0 = 8x^7-12x^3-1
1 votes
Thanks 1
konstantsya
Сделай оценку ответа, пожалуйста)
999Dmitry999
У вас в 7 нет преобразований ,это конечно не считается за ошибкой ,но при нахождении экстремумов нужно будет преобразовывать
999Dmitry999
И там не будет +
999Dmitry999
4x^3++++
999Dmitry999
Нужен минус
999Dmitry999
2 votes
Thanks 0
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Answers & Comments
2) f'(x) = (3x)' = 3
3) f'(x) = (x^(-8))' = -8x^(-9)
4) f'(x) = (3,7)' = 0
5) f'(x) = (1/x)' = (x^(-1))' = -x^(-2) = -1/x^2
6) f'(x) = (3x^8-2x^6+4x^4-3x^2+x-8)' =
= 3*8x^(8-1)-2*6x^(6-1)+4*4x^(4-1)-3*2x^(2-1)+x^0-0 =
= 24x^7-12x^5+16x^3-6x+1
7) 2) f'(x) = (x^4+)' = (x^4+x^(1/2) )' =
= 4x^(4-1)+1/2 *x^(1/2-1) = 4x^3-1/2*x^-1/2
4x^3-
8) f'(x) = (1/4 *x^(-4))' = 1/4*(-4)*x^(-4-1) = -x^(-5)
9) 6) f'(x) = (x^8-3x^4-x+5)' = 8x^(8-1)-3*4x^(4-1)-1X^0+0 = 8x^7-12x^3-1