Question

What is the slope of the tangent line of `1/x^3-y^2-y = C`, where C is an arbitrary constant, at `(-5,0)`?

Answer 1

The slope is `(-3)/625`

Explanation:

The constant `C`
If we know that `(-5,0)` is a point on the graph, then `C = -1/125`.
So `C` is determined by the values of `x` and `y`. It is not arbitrary. (It is not independent of the values of the variables.)

Slope of Tangent
Regardless of that, we can find the slope of the tangent line by implicit differentiation.

`d/dx(1/x^3-y^2-y)=d/dx(C)`

`-3/x^4-2ydy/dx=dy/dx = 0`

`-3/x^4 = (2y+1) dy/dx`

The slope at `(-5,0)`, is therefore

`dy/dx = -3/625`