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)#?

1 Answer
Jun 21, 2016

#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#