What is the equation of the line tangent to the curve xy=4 at (2,2)?

To obtain the slope of this TL at (2,2), differentiate xy=4 with respect to x:

x(dy/dx) + y(dx/dx) = 0.  Then x(dy/dx) = -y, and (dy/dx) = -y/x.

At (2,2), this slope is -2/2, or -1.

Then the eqn of the TL to the curve xy=4 is   y-2 = -1(x-2) = -x+2

Thus, y = 2 - x + 2, or    y = -x + 4  (answer)