AP Calculus question, could someone please explain the below question?

Question

The slope of the line tangent to the graph of ln(xy)=x at the point where x=1 is

1-e

1

0

e^2

e

Answer 1

ln(xy) = x

Using properties of logarithms, lnx + lny = x

So, 1/x + 1/y(dy/dx) = 1.

When x = 1, we have lny = 1. So, y = e.

Therefore, when x = 1, 1 + (1/e)dy/dx = 1.

(1/e)(dy/dx) = 0.

So, since 1/e ≠ 0, dy/dx = 0. In other words, the tangent line to the curve with equation ln(xy) = x at the point (1,e) is horizontal.

Answer 2

There are two ways to find a formula for y' for this function. One is to use implicit differentiation. Another is to first solve for y, then take the derivative with respect to x. Either way the formula discovered gives the slope of the tangent line to the graph of the function at a given point on the curve.

Implicit

1/xy [xy'+y] = 1

xy' + y = xy

xy' = xy - y

y' = (xy - y)/x

To find the value of y' at the point where x = 1 also requires you to determine the value of y where x = 1.

ln(y) = 1

e¹=y

So the point (1, e) is on the curve.

y'(1,e) = [1(e) - e]/1 = 0

Explicit

ln(xy)=x

e^x=xy

y = e^x/x

y' = [xe^x-e^x]/x² [using the quotient rule)

y'(1) = [1e - e]/1 = 0/1 = 0

Visit this graph to see other tangent lines to the curve:

desmos.com/calculator/zhhhrdkl51

Slide point P on the graph to see other tangent lines.