Question

求一阶导数和二阶导数

求y*=xe^x(Acos2x+Bsin2x)的一阶导数和二阶导数详细过程

Answer 1

y = xe^x(Acos2x + Bsin2x) = u(x)v(x)
u(x) = xe^x
u'(x) = xe^x + e^x = (1 + x)e^x
u''(x) = e^x + (1 + x)e^x = (2 + x)e^x
v(x) = Acos2x + Bsin2x
v'(x) = -2Asin2x + 2Bcos2x
v''(x) = -4Acos2x - 4Bsin2x
y' = u'(x)v(x) + u(x)v'(x)
y'' = [u''(x)v(x) + u'(x)v'(x)]+[u'(x)v'(x) + u(x)v''(x)] = u''(x)v(x) + 2u'(x)v'(x) + u(x)v''(x)
将u,v,u',v',u'',v''代入计算就可以得到原函数的一、二阶导数