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)?

Jun 21, 2016

$\frac{1}{2}$

Explanation:

do you mean $\left(x y - \frac{1}{y x}\right) \left(x y - \frac{1}{x y}\right) \setminus \textcolor{red}{=} C$ which is just ${\left(x y - \frac{1}{y x}\right)}^{2} = C$

if so , i'd use the Implicit Function Theorem $\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / {f}_{y}$

using the product rule for the partials we get

${f}_{x} = 2 \left(x y - \frac{1}{x y}\right) \left(y + \frac{1}{{x}^{2} y}\right)$

${f}_{y} = 2 \left(x y - \frac{1}{x y}\right) \left(x + \frac{1}{x {y}^{2}}\right)$

$s o \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 \left(x y - \frac{1}{x y}\right) \left(y + \frac{1}{{x}^{2} y}\right)}{2 \left(x y - \frac{1}{x y}\right) \left(x + \frac{1}{x {y}^{2}}\right)}$

$= - \frac{y + \frac{1}{{x}^{2} y}}{x + \frac{1}{x {y}^{2}}} , \setminus \textcolor{red}{{x}^{2} {y}^{2} \setminus \ne 1}$

$= - \frac{\frac{{x}^{2} {y}^{2} + 1}{{x}^{2} y}}{\frac{{x}^{2} {y}^{2} + 1}{x {y}^{2}}} = - \frac{y}{x}$

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