Question

Find an equation of the tangent to the curve y=xlnx at the point (e,e)

Answer 1

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Answer 2

Find y'. Put x= e, y = e to get the slope at(e,e). Find the equation of the line knowing slope and the point (e,e)

Answer 3

y' = x(1/x) + 1(lnx) = 1 + lnx , so slope of line tangent to the curve at (e,e) is 1 + ln e, or 1 + 1 = 2

Line has a slope of 2 and passes through the point (e,e). Using point-slope form:

y - y1 = m(x - x1) --> y - e = 2(x - e) --> y - e = 2x - 2e --> y = 2x - e

I hope that was helpful!