Question

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)

Answer 1

`(x+y)^3=x^3+y^3`

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

`(\mathrm d(x+y)^3)/(\mathrm dx)=(\mathrm d(x^3+y^3))/(\mathrm dx)`

`3(x+y)^2(\mathrm d(x+y))/(\mathrm dx)=3x^2+3y^2(\mathrm dy)/(\mathrm dx)`

`3(x+y)^2\left(1+(\mathrm dy)/(\mathrm dx)\right)=3x^2+3y^2(\mathrm dy)/(\mathrm dx)`

Solve for
`(\mathrm dy)/(\mathrm dx)`:

`(3(x+y)^2-3y^2)(\mathrm dy)/(\mathrm dx)=3x^2-3(x+y)^2`

`(\mathrm dy)/(\mathrm dx)=(3x^2-3(x+y)^2)/(3(x+y)^2-3y^2)`

`(\mathrm dy)/(\mathrm dx)=(x^2-(x+y)^2)/((x+y)^2-y^2)`

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

`(\mathrm dy)/(\mathrm dx)=((-1)^2-(-1+1)^2)/((-1+1)^2-1^2)=-1`