Determine the tangent line at (-1,1) defined by the equation x^4-x^2y+y^4=1.

Answer 1

We must find the derivative of the function using the concept "implicit differentiation".

Taking the derivative, you will get:

4x^3 -x^2(dy/dx) + y(-2x) + 4y^3(dy/dx) = 0

Getting the dy/dx on one side, you will get:

4y^3(dy/dx) -x^2(dy/dx) = 2xy -4x^3

Factoring dy/dx, you will get:

dy/dx (4y^3 - x^2) = 2xy - 4x^3

Solve for dy/dx, you will get:

dy/dx = (2xy - 4x^3)/(4y^3 -x^2)

dy/dx is the derivative which is the slope, but must find it at given point (-1, 1).

So plug the point in for (x,y)

dy/dx = [2(-1)(1) - 4(-1)^3]/[4(1)^3 - (-1)^2]

= [ -2 + 4]/[4 - 1]

= 2/3

tangent line formula y -y1 = m(x - x1)

y - 1 = 2/3(x + 1)