Question

find the equation of the line normal to the curve defined by the equation x^3y^4-5=x^3-x^2+y at the point (2,-1).

Answer 1

So we start by finding the slope of the line `tangent` to the curve at (2,-1).
From that information we can determine the slope of the line `normal` to the curve.

So to get the slope of the line tangent to the curve, we need to take the derivative.
For this one I guess we'll differentiate `implicitly`.

Need any help with the derivative steps?

Answer 2

I got the slope as 4/33