Question

What is the slope of the tangent line of `(xy-1/(yx))(xy-1/(xy)) C`, where C is an arbitrary constant, at `(-2,1)`?

Answer 1

`1/2`

Explanation:

do you mean `(xy-1/(yx))(xy-1/(xy)) \color{red}{=} C` which is just `(xy-1/(yx))^2= C`

if so , i'd use the Implicit Function Theorem `dy/dx = - f_x/f_y`

using the product rule for the partials we get

`f_x = 2(xy- 1/(xy)) (y + 1 / (x^2y))`

`f_y = 2( xy- 1/(xy)) (x + 1 / (xy^2))`

`so dy/dx =- (2(xy- 1/(xy)) (y + 1 / (x^2y))) / (2( xy- 1/(xy)) (x + 1 / (xy^2)) )`

`=- (y + 1 / (x^2y)) / (x + 1 / (xy^2)) , \color{red}{x^2 y^2 \ne 1}`

`=- ((x^2y^2 + 1)/(x^2 y) ) / ((x^2y^2 + 1)/(xy^2) ) = -y/x`

so, `dy/dx ]_{(-2,1)} = 1/2`