Find the equation of the tangent to the curve x^3-2y-8x-3=0

Answer 1

Hi Albert

Given x³-2y-8x=0 to find the equation of the tangent (dy/dx) you can differentiate both sides

3x²-2dy/dx-8=0 now we solve for dy/dx

dy/dx=(3/2)x²-4 which is the equation of the tangent.

Does this help?

Jim