Regard as the independent variable and as the dependent variable and find the slope of the tangent line to the curve (x^2 + y^2)^2 - 4x^2 y = 225 at the point (x,y) = (4,1)

Question

Thank you very much for helping me.

Answer 1

A little implicit differentiation helps out here.

2(x^2 + y^2)(2x + 2yy') - (8xy + 4x^2 y') = 0
4x(x^2 + y^2) + 4y(x^2 + y^2) y' = 8xy + 4x^2 y'

y'(4x(x^2 + y^2) - 4x^2) = 8xy - 4x(x^2 + y^2)

y' = 4x(2y - (x^2 + y^2))/(4x^3 - 4x^2 + 4xy^2)
at (4,1) we have
y' = 4*4(2-17)/(256 - 64 + 16)
y' = -16*15/208 = -1.15

Better double-check the algebra.