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fb17
@fb17
July 2022
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Легко, помогите
Найдите две производных g(x)=3x-(1/x³) и g(x)=1/2cosx
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m11m
Verified answer
1)
g(x)=3x - (1/x³)=3x - x⁻³
g ' (x)=3 - (-3)x⁻⁴ = 3 + (3/x⁴)
2) Не понятно условие.
1 случай:
g(x)=1/2cosx=0.5cosx
g ' (x)= -0.5sinx
2 случай:
g(x)= 1 / (2cosx) (в числителе - 1; в знаменателе - 2cosx) =
= (1/2) (cosx)⁻¹
g ' (x)=(1/2) * (- 1/cos²x) * (-sinx) = sinx / (2cos²x)
1 votes
Thanks 1
nKrynka
Решение
1) g(x)=3x-(1/x³) = 3x - x
⁻³
g`(x) = 3 - 3/x
⁴
2) g(x)=1/2cosx = (1/2)*cos
⁻¹x
g`(x) = (1/2)*(- 1)*cos⁻¹⁻¹x * (- sinx) = sinx/[2*cos²x)
0 votes
Thanks 1
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Answers & Comments
Verified answer
1)g(x)=3x - (1/x³)=3x - x⁻³
g ' (x)=3 - (-3)x⁻⁴ = 3 + (3/x⁴)
2) Не понятно условие.
1 случай:
g(x)=1/2cosx=0.5cosx
g ' (x)= -0.5sinx
2 случай:
g(x)= 1 / (2cosx) (в числителе - 1; в знаменателе - 2cosx) =
= (1/2) (cosx)⁻¹
g ' (x)=(1/2) * (- 1/cos²x) * (-sinx) = sinx / (2cos²x)
1) g(x)=3x-(1/x³) = 3x - x⁻³
g`(x) = 3 - 3/x⁴
2) g(x)=1/2cosx = (1/2)*cos⁻¹x
g`(x) = (1/2)*(- 1)*cos⁻¹⁻¹x * (- sinx) = sinx/[2*cos²x)