Calculus derivatives and tangent line

Question

Consider the curve x^2y + y^2x = 6

A) Find dy/dx in terms of x and y

B) Write the equation for the tangent lien where x =2 and y=1

C) Find the coordinates of the point (x,y) where the curve has a horizontal tangent line.

Answer 1

x²y + y²x = 6

a)d/dx(x²y + y²x) = d/dx(6); 2xy + x²y' + y² + 2xyy' = 0; y'(x² + 2xy) = - (2xy + y²);

y' = - y(2x + y)/[x(x + 2y)]

b) At x = 2 y = 1 and y' = - 1( 4 + 1)/[2(2 + 2)] = - 5/8.

So, tangent line: y = 1 - 5/8(x - 2); y = - 5/8x + 9/4;

c) y' = 0; y(2x + y) = 0; y = 0 don't satisfy given line equation (0 = 6 contradiction).So, 2x + y = 0; y = - 2x;

1. 2x³ + 4x³ = 6; x³ = 3; x = 3^1/3 and y = -·2 3^1/3
2. At point ( 3^1/3, - 2·3^1/3) graph has horizontal tangent line