I really need help with this derivatives problem (calculus 1). (chegg tutors have given me very broad explanations so I've been stuck on this one for hours)

dy/dx = (1 − 8xy + 2y² + 16x⁴y² + 32x⁴y⁴) / (4x² − 4xy − 64x⁵y³)

Step-by-step explanation:

tan⁻¹(4x²y) = x + 2xy²

Take derivative of both sides with respect to x.  You'll need to use the following derivatives and rules:

Derivative of tan⁻¹: d/dx (tan⁻¹ u) = 1 / (1 + u²) du/dx

Product rule: d/dx (uv) = u dv/dx + v du/dx

Power rule: d/dx (u^n) = n u^(n−1) du/dx

(Notice the chain rule in each.)

So the derivative of the left side is:

d/dx [tan⁻¹(4x²y)] = 1 / (1 + (4x²y)²) × (4 (x² dy/dx + 2xy))

d/dx [tan⁻¹(4x²y)] = (4x² dy/dx + 8xy) / (1 + 16x⁴y²)

And the derivative of the right side is:

d/dx (x + 2xy²) = 1 + 2 (2xy dy/dx + y²)

d/dx (x + 2xy²) = 1 + 4xy dy/dx + 2y²

Set the sides equal and solve for dy/dx.

(4x² dy/dx + 8xy) / (1 + 16x⁴y²) = 1 + 4xy dy/dx + 2y²

4x² dy/dx + 8xy = (1 + 4xy dy/dx + 2y²) (1 + 16x⁴y²)

4x² dy/dx + 8xy = (1 + 2y²) (1 + 16x⁴y²) + 4xy (1 + 16x⁴y²) dy/dx

4x² dy/dx + 8xy = (1 + 2y² + 16x⁴y² + 32x⁴y⁴) + (4xy + 64x⁵y³) dy/dx

(4x² − 4xy − 64x⁵y³) dy/dx = 1 − 8xy + 2y² + 16x⁴y² + 32x⁴y⁴

dy/dx = (1 − 8xy + 2y² + 16x⁴y² + 32x⁴y⁴) / (4x² − 4xy − 64x⁵y³)

g(x) is the negative form of f(x)? because whenever its the negative form it reflects its self over the x axis. idk if this is right though.

step-by-step explanation:

i belive those are the ones that are under 90 degreese.