Find dy/dx implicit differentiation of each equation and evaluate at the given point to find the slope at that point.

(x+y)3=x3+y3 at (-1,1)

Differentiate both sides with respect to . By the power and chain rules, we have

Solve for :

At the point (-1, 1), the slope of the tangent line to the curve is

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)