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

Apr 23, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 1.5$

#### Explanation:

We first find $\frac{d}{\mathrm{dx}}$ of each term.

$\frac{d}{\mathrm{dx}} \left[x {y}^{2}\right] - \frac{d}{\mathrm{dx}} \left[{\left(1 - x y\right)}^{2}\right] = \frac{d}{\mathrm{dx}} \left[C\right]$

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

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

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

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

The chain rule tells us:
$\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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

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

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

For $\left(1 , - 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{\left(- 1\right)}^{2} + 2 \left(- 1\right) \left(1 - 1 \left(- 1\right)\right)}{2 \left(1\right) \left(- 1\right) - 2 \left(1\right) \left(1 - 1\right)} = - 1.5$