# What is the slope of the tangent line of xy^2-(1-x/y)^2= C , where C is an arbitrary constant, at (1,-1)?

Nov 4, 2016

slope=$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2}$

#### Explanation:

To find the slope of the tangent line you have to find the derivative first then plug in the point for x and y

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

${y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 \left(1 - \frac{x}{y}\right) \cdot \left(- \frac{1}{y}\right) - 2 \left(1 - \frac{x}{y}\right) \left(\frac{x}{y} ^ 2\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

${y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{2 y - 2 x}{y} ^ 2 + \frac{2 {x}^{2} - 2 x y}{y} ^ 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{2 {x}^{2} - 2 x y}{y} ^ 3 \frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{2} - \frac{2 y - 2 x}{y} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 x y + \frac{2 {x}^{2} - 2 x y}{y} ^ 3\right) = - {y}^{2} - \frac{2 y - 2 x}{y} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{2 x {y}^{4} + 2 {x}^{2} - 2 x y}{y} ^ 3\right) = \frac{- {y}^{4} - 2 y + 2 x}{y} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {y}^{4} - 2 y + 2 x}{y} ^ 2 \cdot {y}^{3} / \left(2 x {y}^{4} + 2 {x}^{2} - 2 x y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {y}^{5} - 2 {y}^{2} + 2 x y}{2 x {y}^{4} + 2 {x}^{2} - 2 x y}$

dy/dx=(-(-1)^5-2(-1)^2+2(1)(-1))/(2(1)(-1)^4+2(1)^2-2(1)(-1)

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{6} = - \frac{1}{2}$