# What is the slope of the tangent line of  (x^2-y^2)/(sqrt(4y-x)+xy) =C , where C is an arbitrary constant, at (1,1)?

Mar 17, 2018

#### Explanation:

Given that (1,1) lies on the graph, we see that C = 0#.

The equation is equivalent to ${x}^{2} - {y}^{2} = 0$ with $y \ge 0$.

Differentiating implicitly, we have:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x} = \left\{\begin{matrix}1 & \text{if" & x > 0 \\ -1 & "if} & x < 0\end{matrix}\right.$

At $\left(1 , 1\right)$, the slope is $1$.

$\frac{{x}^{2} - {y}^{2}}{\sqrt{4 y - x} + x y} = 0$ is graphed below.

(Note that ${x}^{2} - {y}^{2} = \left(x - y\right) \left(x + y\right)$ so the graph is parts of $y = x$ and $y = - x$.)

graph{(x^2-y^2)/(sqrt(4y-x)+xy) =0 [-6.496, 7.55, -1.824, 5.2]}