Question

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

Answer 1

The slope is `0`

Explanation:

`5x^3y^2-y = C` at `(0,0)`

`15x^2y^2+10x^3ydy/dx - 1 dy/dx = 0`

At `(0,0)`, we get

`0+0-1dy/dx=0`, so `dy/dx=0`