y^3x + y^2x^2 = 6
Using implicit differentiation
3 y^2.y'.x+y^3+2 y.y'.x^2+2 y^2.x=0
y'(3 y^2.x+2 y. x^2) = -y^3-2 y^2.x
y' = -[y^2(y+2 x)]/[x y(3 y+2 x)]
= -[y(y+2 x)]/[x(3 y+2 x)]
Substituting x =2 and y = 1:
y' = -(1)(1+4)/[2(3+4)]
= -5/14
The slope of the tangent of the curve at (2,1) will be:
m = -5/14
