微积分 还有几道啊，恩人，能否把求解的过程写详细一点，多谢多谢

Question

如题所述

Answer 1

1.取对数得：lny=1/2(ln(x^2-8)+ln(x^3+6)-ln(x-3)-ln(x^2-4))
两边求到得：y'/y=1/2*(2x/(x^2-8)+3x/(3x^2+6)-1/(x-3)-2x/(x^2-4))
y'=1/2*y*(2x/(x^2-8)+3x/(3x^2+6)-1/(x-3)-2x/(x^2-4)) 自己把y代入
2.y=incose^(-xx)，求到得：y'=-sine^(-xx)*(e^(-xx))*(-2x)/cose(-xx)
=2x*e^(-x^2)*tane^(-x^2)
3.xy=e^(x+y)，两边求导得：y+xy'=e^(x+y)*(1+y')
y-e^(x+y)=(e^(x+y)-x)y'，所以：y'=( y-e^(x+y))/(e^(x+y)-x)=(y-xy)/(xy-x)
5.lim(x^2/(1-√(1+x^2))=lim(x^2*(1+√(1+x^2))/((1-√(1+x^2)((1+√(1+x^2))
=lim(x^2(1+√(1+x^2))/(1-1-x^2)=lim - ((1+√(1+x^2))= -2来自：求助得到的回答