How do you find the equation of the tangent line to the curve $y=x-sqrtx$ at (1,0)?

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Answer 1 · 2017-01-04T11:50:04

Just to welcome a newcomer, with an illustrative graph, for the nice answer,

To make y real, $x >=0$ .

The inserted Socratic graph is tangent-inclusive.

graph{(2y-x+1)(y-x+sqrtx)=0x^2 [-2.5, 2.5, -1.25, 1.25]}

How do you find the equation of the tangent line to the curve y=x-sqrtxy=x-sqrtx at (1,0)?