Find the slope of the tangent to x^2 +y^3 =5 at the point where x = 2

Answer 1

taking the derivative, we have
2x + 3y^2 y' = 0
y' = -(2x)/(3y^2)
so, now we need to find y when x=2:
4+y^3 = 5
y = 1
So, at (2,1), the tangent has slope -4/3