函数f(x)=sinx+cosx在点(0,f(0))处的切线方程为()A.x-y+1=0B.x-y-1=0C.x+y-1=0D.x+y+1=0-数学

题目简介

函数f(x)=sinx+cosx在点(0,f(0))处的切线方程为()A.x-y+1=0B.x-y-1=0C.x+y-1=0D.x+y+1=0-数学

题目详情

函数f(x)=sinx+cosx在点(0,f(0))处的切线方程为(  )
A.x-y+1=0B.x-y-1=0C.x+y-1=0D.x+y+1=0
题型:单选题难度:中档来源:不详

答案

∵f(x)=sinx+cosx
∴f′(x)=cosx-sinx
∴f'(0)=1,所以函数f(x)在点(0,f(0))处的切线斜率为1;
又f(0)=1,
∴函数f(x)=sinx+cosx在点(0,f(0))处的切线方程为:
y-1=x-0.即x-y+1=0.
故选A.

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