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 )?

1 Answer
Jun 26, 2015

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#.