Answer:
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)
