student grade 12 calculus
write an equation of the line tangent to the graph of

ey + ln(xy) = 1 + e at (e,1)

please help cant figure out


I first checked to see if (e,1) satisfies the equation using substitution

e1 + ln(e) = e + 1 Hence (e,1) is on the graph. Thus the tangent line is the line through (e,1) with slope given by the derivative of the expression at the point (e,1). You can find the derivative using implicit differentiation.

Differentiating both sides of ey + ln(xy) = 1 + e I get

Can you complete the problem now?
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