Find an equation of the tangent line to the curve at the point corresponding to the given value of the parameter: x= e^square root of t; y=t- 2lnt; t=1

tangent-line

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Find an equation of the tangent line to the curve at the point corresponding to the given value of the parameter: x= e^square root of t; y=t- 2lnt; t=1.

Answer:

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