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

1 Answer
Eddie
Jun 21, 2016

1/212

Explanation:

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

if so , i'd use the Implicit Function Theorem dy/dx = - f_x/f_ydydx=−fxfy

using the product rule for the partials we get

f_x = 2(xy- 1/(xy)) (y + 1 / (x^2y)) fx=2(xy−1xy)(y+1x2y)

f_y = 2( xy- 1/(xy)) (x + 1 / (xy^2)) fy=2(xy−1xy)(x+1xy2)

so dy/dx =- (2(xy- 1/(xy)) (y + 1 / (x^2y))) / (2( xy- 1/(xy)) (x + 1 / (xy^2)) ) sodydx=−2(xy−1xy)(y+1x2y)2(xy−1xy)(x+1xy2)

=- (y + 1 / (x^2y)) / (x + 1 / (xy^2)) , \color{red}{x^2 y^2 \ne 1}=−y+1x2yx+1xy2,x2y2≠1

=- ((x^2y^2 + 1)/(x^2 y) ) / ((x^2y^2 + 1)/(xy^2) ) = -y/x=−x2y2+1x2yx2y2+1xy2=−yx

so, dy/dx ]_{(-2,1)} = 1/2dydx](−2,1)=12

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