题目内容
已知函数f(x)=+xln x,则曲线y=f(x)在x=1处的切线方程为( )
A.x-y-3=0 | B.x-y+3=0 | C.x+y-3=0 | D.x+y+3=0 |
C
f′(x)=-+ln x+1,当x=1时f′(1)=-1.f(1)=2,即切点坐标为(1,2).故切线方程为y-2=-(x-1),即x+y-3=0.
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题目内容
A.x-y-3=0 | B.x-y+3=0 | C.x+y-3=0 | D.x+y+3=0 |