Expert: Paul Klarreich - 2/22/2006

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

x^2/3 + x^2/3 = 4   given point: (-3*sqrt3,1)  

Answer
Hi, chevonne,

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

x^2/3 + x^2/3 = 4 given point: (-3*sqrt(3),1)

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I assume you meant:

x^2/3 + y^2/3 = 4 given point: (- 3*sqrt(3),1)

This is mostly the same as the last one, but you don't need the second derivative.

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

 x^(-1/3) + y^(-1/3) y' = 0

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

Now you want the slope.  Use your  x = -3*sqrt(3),  y = 1

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

Now  -3^1/3 is - (3^1/3)  and  sqrt(3)^1/3 = 3^1/6, so the denominator is  - (3)^(1/3 + 1/6) = - 3^1/2 = - sqrt(3)

         - 1        sqrt(3)
So  m = --------- =  -------
       - sqrt(3)       3

OK, we're ready to write the line, using the slope and the point:
       sqrt(3)
y - 1 = -------(x + 3 sqrt(3))
          3
       sqrt(3)
y - 1 = ------- x + 3
          3

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