X^2/3 + Y^2/3 = 4 (-3square root of 3, 1)?

Question

use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Answer 1

2/3x^(-2/3) +y' 2/3y^-2/3 =0
y' = [-2/3x^(-2/3)]/ 2/3y^(-2/3)
y' = (-y^2/3)/x^2/3
y' = -1/(-3sqrt(3))^2/3 = 1/3
y = x/3+b
1 = -3*rt(3)/3 + b
b = 1+sqrt(3)
y = x/3 +1+sqrt(3)

Answer 2

x^2/3 + y^2/3 = 4

differentiating

2/3 x^(-1/3) + 2/3y^(-1/3) y' = 0

x^(-1/3) = - y^(-1/3)y'

y' = -(y/x)^1/3)

point of tangency = -3sqrt(3), 1

slope = -1/(-3sqrt(3)^1/3

= 1/sqrt(3)

equation of tangent

y-y1 = m(x-x1)

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

sqrt(3)[y-1] = x + 3 sqrt(3)

y sqrt(3) - sqrt(3) = x + 3sqrt(3)

y sqrt(3) = x +4 sqrt(3)

divide by sqrt(3)

y =[ x/sqrt(3)] + 4