Question 683247
Rewrite the equation to isolate y on the LHS:
y^(2/3) = 4 - x^(2/3)
y = (4 - x^(2/3))^(3/2)
The slope of the tangent line is the derivative, dy/dx at the point x=-1:
dy/dx(x=1) = (3/2)(4-x^(2/3))^(1/2)*(-2/3)(x^(-1/3) = -(4-x^(2/3))^(1/2)/x^(1/3) = -(4-1)^1/2/-1 = sqrt(3) = 1.7321