[tex](x+y)^3=x^3+y^3[/tex]
Differentiate both sides with respect to [tex]x[/tex]. By the power and chain rules, we have
[tex]\dfrac{\mathrm d(x+y)^3}{\mathrm dx}=\dfrac{\mathrm d(x^3+y^3)}{\mathrm dx}[/tex]
[tex]3(x+y)^2\dfrac{\mathrm d(x+y)}{\mathrm dx}=3x^2+3y^2\dfrac{\mathrm dy}{\mathrm dx}[/tex]
[tex]3(x+y)^2\left(1+\dfrac{\mathrm dy}{\mathrm dx}\right)=3x^2+3y^2\dfrac{\mathrm dy}{\mathrm dx}[/tex]
Solve for [tex]\frac{\mathrm dy}{\mathrm dx}[/tex]:
[tex](3(x+y)^2-3y^2)\dfrac{\mathrm dy}{\mathrm dx}=3x^2-3(x+y)^2[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2-3(x+y)^2}{3(x+y)^2-3y^2}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{x^2-(x+y)^2}{(x+y)^2-y^2}[/tex]
At the point (-1, 1), the slope of the tangent line to the curve is
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(-1)^2-(-1+1)^2}{(-1+1)^2-1^2}=-1[/tex]
