Question

How do you use implicit differentiation to find the equation of the tangent line to the curve `xy^3+xy= 2` at the point ( 1 , 1 )?

Answer 1

Use implicit differentiation to find `dy/dx`, eveluate `d/dx` at `(1/1)` to get the slope of the tangent line. Find the equation of the line.

Explanation:

`xy^3+xy= 2`

`d/dx(xy^3+xy) = d/dx(2)`

`d/dx(xy^3)+d/dx(xy)= d/dx(2)`

Both terms on the left are products, so we'll use the product rule. I use the order: `d/dx(uv) = u'v+uv'`

`underbrace[(1)y^3 +x(3y^2 dy/dx)] _ (d/dx(xy^3))+ underbrace[(1)y +x(dy/dx)] _ (d/dx(xy)) = underbrace(0) _ (d/dx(2))`

Simplify to get:

`y^3+3xy^2 dy/dx +y +x dy/dx =0`
Solve for `dy/dx`:

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

At `(1,1)`, the slope of the tangent line is `=1/2` and the equation of the line is:

`y=-1/2x+3/2`.