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15 лет назад
y'=1/2* 1/[ ln(sin 2x^2+2)^1/2] * 1/ [sin 2x^2+2] * (sin 2x^2+2)' = 1/2* 1/ I (sin 2x^2+2)*[ ln(sin 2x^2+2)^1/2] I * 4xcos 2x^2 =
= 2xcos 2x^2 / { (sin 2x^2+2)*[ ln(sin 2x^2+2)^1/2] }
2.
y'=e^ln(x^2-4)* { ln(x^2-4) } ' = 2xe^ln(x^2-4) / (x^2-4)
y=e^ln(x^2-4)