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

Feb 2, 2016

The slope of the tangent is $0$ (and $C$ is not "arbitrary", it is $- 3$).

#### Explanation:

$\left(x - y\right) \left(x + y\right) - x y + y = C$

Instead of differentiating immediately, let's rewrite the first product.

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

This form is (I think) simpler to differentiate.

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

At the point $\left(1 , 2\right)$, we get

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

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ to see that $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$