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)