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

Mar 6, 2016

It is $\frac{1}{2}$

#### Explanation:

Given that $\left(1 , 1\right)$ lies on the graph of $\frac{x - {y}^{2}}{x {e}^{x - {y}^{2}}} = C$,

we see that $C = 0$

(Substitute $1$ for both $x$ and $y$ to get
(1-(1)^2)/(1e^(1-(1^2))=0/1=0=C.)

That means the graph is the graph of $x - {y}^{2} = 0$ or ${y}^{2} = x$.

Now we can differentiate implicitly, of further recognize that,

since $\left(1 , 1\right)$ is on the graph, we are on the branch where $y = \sqrt{x}$, so $y ' = \frac{1}{2 \sqrt{x}}$

The slope at $x = 1$ is $\frac{1}{2}$.