Implicit Differentiation to find equation of a tangent line

[hannahSUU]

I need urgent help. I have this question:
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
\begin{equation}
{x}^{2/3}+{y}^{2/3}=4
\\
\left(-3\sqrt{3}, 1\right)\end{equation}

(astroid)
[DESMOS=-10,10,-10,10]x^{\frac{2}{3}}+y^{\frac{2}{3}}=4[/DESMOS]

My answer is
\begin{equation}
-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}\left(x+3\sqrt{3}\right)+1
\end{equation}
which is marked as incorrect, but I can't figure out why.

Any help is appreciated (heart)

EDIT: I clearly do not know how to use Latex. Working on fixing it now
EDIT 2: Fixed

 

[topsquark]

Look at your function again.
\(\displaystyle x^{2/3} + y^{2/3} = 4\)

Take your derivative.
\(\displaystyle \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} ~ y' = 0\)

Now solve for y'.

-Dan

 

[HallsofIvy]

I need urgent help. I have this question:
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
\begin{equation}
{x}^{2/3}+{y}^{2/3}=4
\\
\left(-3\sqrt{3}, 1\right)\end{equation}

(astroid)My answer is
\begin{equation}
-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}\left(x+3\sqrt{3}\right)+1
\end{equation}
which is marked as incorrect, but I can't figure out why.

Any help is appreciated (heart)

EDIT: I clearly do not know how to use Latex. Working on fixing it now
EDIT 2: Fixed

I thought I had responded to this before. Perhaps it was on another math site?

One reason this was marked as incorrect was that the problem asked for "the equation of the tangent line" and what you have is not an equation! There is no "=".

But even if you intended "y= " in front of that, it is not the equation of a line. You need to evaluate that derivative at the given point, not just leave it in terms of "x" and "y".

From $$x^{2/3}+ y^{2/3}= 4$$, $$(2/3)x^{-1/3}+ (2/3)y^{-1/3}y'= 0$$ so $$y'= -\frac{y^{1/3}}{x^{1/3}}$$. Now, specifically, at $$(-3\sqrt{3}, 1)$$ that becomes $$y'(-3\sqrt{3})= -\frac{1^{1/3}}{(-3\sqrt{3})^{1/3}}$$. Now, the tangent line is given by $$y- 1= \left(-\frac{1^{1/3}}{(-3\sqrt{3})^{1/3}}\right)(x- 3\sqrt{3})= 3^{-1/2}(x- 3\sqrt{3})$$.

 

[Greg]

That should be $ y - 1 = 3^{-1/2}(x + 3\sqrt3)$ :)

 

[HallsofIvy]

Thanks.