e是常數(shù),f'(x)=1/x,所以切線在x=1點的斜率是1,在x=1點的函數(shù)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?
數(shù)學人氣：461 ℃時間：2020-02-05 18:31:47
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