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

Answer 1

I agree with Paul.

Shailesh K, Senior Math Instructor

Answer 2

You haven't given the point, but I will tell you what to do when yo have it.

The equation becomes 2y = x³ - 8x - 3

Take the derivative on both sides

2 (dy/dx) = 3x² - 8 => dy/dx = (1/2)(3x² - 8)

Plug in the point and you you have the slope of the tangent, then use the point-slope method to define the equation of the tangent line.