Find the equation of the tangent line to the curve ln(xy)=2x at the point (1,e^2).

Question

Tangent line y=______________

Answer 1

ln(xy) = 2x find slope at (1,e^2)

e^lnxy = e^2x

xy = e^2x

(xy)' =xy' +y = 2e^2x

xy' = 2e^2x - y

y' = (2e^2x-y)/x

y'(1,e^2) = (2e^2 -e^2)/1 = e^2

tangent line has slope = e^2

and goes through the point (1,e^2)

y-e^2 = e^2(x-1) = xe^2 -e^2

y = xe^2 is the tangent line

or

ln(xy) = 2x

(1/xy)(xy'+y)= y'/y + 1/x= 2

y'/y = 2 - 1/x

y' = (2-1/x)(y)

y'(1,e^2) = e^2

y-e^2 = e^2(x-1) = xe^2 - e^2x

y = xe^2 is the tangent line through the point (1, e^2) to the curve lnxy=2x