求微分两题y=(e^(-2x))*Cos(3+2x)y=x/(1-x^2)

Question

求微分两题
y = (e^(-2x))*Cos(3+2x)
y = x/(1-x^2)

Answer 1

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)
[f(p(x))]'=f'((p(x))p'(x)
y = (e^(-2x))*Cos(3+2x)
y'=(e^(-2x)*(-2x)'cos(3+2x)+e^(-2x)(-sin(3+2x))(3+2x)'
=-2e^(-2x)cos(3+2x)-2sin(3+2x)e^(-2x)
所以
dy=[-2e^(-2x)cos(3+2x)-2sin(3+2x)e^(-2x)]dx
y = x/(1-x^2)
y'=[(1-x^2)-x(1-x^2)']/(1-x^2)^2
=[1-x^2+2x^2]/(1-x^2)^2
=(x^2+1)/(1-x^2)^2
所以
dy=(x^2+1)/(1-x^2)^2dx
注:
求微分后面别忘记加dx