方程y=Inx 再点(e.1)处的切线
问题描述:
方程y=Inx 再点(e.1)处的切线
答
x = e,y = 1
y' = 1/x
k = y'|(x=e) = 1/e
y = x/e + b
x = e ,y = 1 + b ,b=0
切线为: y = x/e
答
答:y=1/e)x
因为,y'=1/x.则(e.1)处的切线线的斜率:1/e
切线为y-1=(1/e)(x-e)
故切线为y=1+(1/e)(x-e) ,即y=(1/e)x
答
求导
=1/e,所以斜率是1/e
y=(1/e)x+b
将点(e,1)代入
b=0
所以方程为y=(1/e)x