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Find an Equation of the Tangent Line to the Curve x=etx = e ^ { \sqrt { t } }

Question 8

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Find an equation of the tangent line to the curve at the point corresponding to the value of the parameter. x=etx = e ^ { \sqrt { t } } , y=tlnt6y = t - \ln t ^ { 6 } ; t=1t = 1

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