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Alekseyiz
@Alekseyiz
August 2022
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Найти производную функции а) y=2e^-x + x^3 б) y=3^x + 3/x в)y=ln x/2 - e^x г)e^3 -8log5(5-основание) X
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skyne8
Verified answer
A) y'=(2e^-x)'+(x^3)'= -2e^-x +3x^2
б) y' = (3^x)' + (3/x)'= (e^(xln3))' + (3/x)'=(ln3)*3^x - 3/x^2
в) y' = (ln x/2)'- (e^x)' = (x/2)'/(x/2) - e^x= (1/x) - e^x
г) y' = (e^3)' - (8log(5)x)' = 0 - 8/(ln5*lnx) = -8/(ln5*lnx),
ln - натуральный логарифм
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Answers & Comments
Verified answer
A) y'=(2e^-x)'+(x^3)'= -2e^-x +3x^2б) y' = (3^x)' + (3/x)'= (e^(xln3))' + (3/x)'=(ln3)*3^x - 3/x^2
в) y' = (ln x/2)'- (e^x)' = (x/2)'/(x/2) - e^x= (1/x) - e^x
г) y' = (e^3)' - (8log(5)x)' = 0 - 8/(ln5*lnx) = -8/(ln5*lnx),
ln - натуральный логарифм