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

Jul 22, 2016

I get $m = \frac{1}{4}$

#### Explanation:

Given that $x$ and $y$ are related by $\left(x - y\right) \left(x + y\right) - x y + y = C$ and that the point $\left(- 1 , - 1\right)$ lies on the graph of the equation, we could find $C$ and the solve for $y$ and differentiate.

That looks like a lot of work. It will probably be easier to differentiate implicitly.

Differentiating the expression $\left(x - y\right) \left(x + y\right)$ involves the product rule and that is needlessly complicated. We can expand this product of a sum and difference to get

${x}^{2} - {y}^{2} - x y + y = C$.

Differentiating, (using the product rule for $- x y$) we get

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

Rather than solve for $\mathrm{dy} ' \mathrm{dx}$ before substituting, I find it simpler to substitue the values $x = - 1$ and $y = - 1$, then solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

$- 2 + 2 \frac{\mathrm{dy}}{\mathrm{dx}} + 1 + \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$- 1 + 4 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{4}$