Find dy/dx by implicit differentiation :

Question

tan ^-1 (yx^2) = x +xy^2

 

 

p215, #17

Answer 1

tan^-1 (x²y) = x + xy²

 

[1/(1+(x²y)²)] (2xy + x² dy/dx) = 1 + y² + 2xy dy/dx

 

2xy/(1+x⁴y²) + [x²/(1+x⁴y²)] dy/dx = 1 + y² + 2xy dy/dx

 

[x²/(1+x⁴y²) - 2xy] dy/dx = 1 + y² + x²/(1+x⁴y²)

 

[x² - 2xy(1+x⁴y²)] dy/dx = (1+y²)(1+x⁴y²) + x²

 

(x² - 2xy - 2x⁵y³) dy/dx = 1 + x² + y² + x⁴y² + x⁴y⁴

 

dy/dx = (1 + x² + y² + x⁴y² + x⁴y⁴) / (x² - 2xy - 2x⁵y³)