Question

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)`?

Answer 1

It is `1/2`

Explanation:

Given that `(1,1)` lies on the graph of `(x-y^2)/(xe^(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 `(1,1)` is on the graph, we are on the branch where `y=sqrtx`, so `y'=1/(2sqrtx)`

The slope at `x=1` is `1/2`.