Thread: Horizontal Tangent - Implicit Differentiation

a curve in the plane is given by the following equation
x^4+2x^2y+3y^2=6

A) Find an expression for dy/dx using implicit differentiation
B) Find the four points on the curve where the tangent line is horizontal

Ok I know how to solve for the derivative. I got
dy/dx = (4x^3-4xy)/(2x^2+6y)

However Im not quite sure how to solve when the tangent is horizontal. I that the tangent is horizontal when the derivative equals 0. However I don't know how solve for this when it is an implicit derivative.

Originally Posted by rawkstar

a curve in the plane is given by the following equation
x^4+2x^2y+3y^2=6

A) Find an expression for dy/dx using implicit differentiation
B) Find the four points on the curve where the tangent line is horizontal

Ok I know how to solve for the derivative. I got
dy/dx = (4x^3-4xy)/(2x^2+6y)

However Im not quite sure how to solve when the tangent is horizontal. I that the tangent is horizontal when the derivative equals 0. However I don't know how solve for this when it is an implicit derivative.

I am assuming your derivative is correct ...

if



,

sub and find the corresponding y-value(s)

sub and find the corresponding x-values

sub it into the original equation?

Originally Posted by rawkstar

sub it into the original equation?

yes ... you are trying to find points on the curve.

ok so i subbed those into the orignial equation and got two x values and two y values
do i then plug those into the original to get the corresponding opposite value

Originally Posted by rawkstar

ok so i subbed those into the orignial equation and got two x values and two y values
do i then plug those into the original to get the corresponding opposite value

when you sub in x = 0 , you get two values for y ... the two points will be and

when you sub in for , you get two values for y ... square those to get the corresponding x-values.