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Pandoritta
@Pandoritta
July 2022
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Решите уравнение xf'(x)=2f(x) если f(x)=(x^3)lnx
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ShirokovP
Verified answer
Найдем производную
f' (x) = (x^3 lnx)' = 3x^2lnx + x^2
x* (3x^2 lnx + x^2) = 2x^3lnx
3x^3lnx + x^3 - 2x^3lnx = 0
x^3lnx + x^3 = 0
x^3 (lnx + 1) = 0
x^3 = 0 ==> x = 0;
lnx = - 1 ==> x = 1/e
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Answers & Comments
Verified answer
Найдем производнуюf' (x) = (x^3 lnx)' = 3x^2lnx + x^2
x* (3x^2 lnx + x^2) = 2x^3lnx
3x^3lnx + x^3 - 2x^3lnx = 0
x^3lnx + x^3 = 0
x^3 (lnx + 1) = 0
x^3 = 0 ==> x = 0;
lnx = - 1 ==> x = 1/e