第1个回答 推荐于2016-12-01
1)y=ln(x2+3)
y'=1/(x²+2)*(x²+3)'=2x/(x²+3)
2)y=e^xcosx
y'=e^xcosx+e^x(-sinx)
=e^x(cosx-sinx)
3) y=√x-1/x
y'=1/(2√x)+1/x²
是这样吧
y=√[(x-1)/x ]
y'=1/2*√[x/(x-1)]*[(x-1)/x]'
=1/2*√[x/(x-1)]*[x-(x-1)])/x²
=1/2*√[x/(x-1)]*1/x²
=1/(2x²)*√[x/(x-1)]
4)Y=x²/(x-1 )
y'=[2x(x-1)-x²]/(x-1)²
=(x²-2x)/(x-1)²本回答被提问者和网友采纳