Question

Find an equation of the Tangent Line to the equation x^(2/3)+y^(2/3) = 4 at the point (-3sqrt(3), 1)

the exponents in this really tripped me up!!! I ended up with the slope being 1/9^(1/18) I think this is wrong? can someone PLEASE help, I will award a medal! :)
Please show how to solve this

Answer 1

implicit .... works great

Answer 2

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

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

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

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

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


or simply: y' = - cbrt(y/x)

Answer 3

-3sqrt(3), 1

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

Answer 4

\[-\frac{(-1)^{1/3}}{(3(3)^{1/2})^{1/3}}\]
\[-\frac{-1}{((27)^{1/2})^{1/3}}\]
\[\frac{1}{(27)^{1/6}}=\frac{1}{3^{1/2}}\]

Answer 5

27 = 3^3 ... so 3^3/6 = 3^1/2

Answer 6

Thanks! I ultimately came up with (sqrt(3)/3)x +4