e是常数,f'(x)=1/x,所以切线在x=1点的斜率是1,在x=1点的函数值是e.
So the equation of the tangent line of graph at x=1 is f(x)=x+e-1
Find the equation of line tangent to the graph of f(x)=e+ln x at x=1?
Find the equation of line tangent to the graph of f(x)=e+ln x at x=1?
数学人气：830 ℃时间：2020-02-05 18:31:47
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