# Thread: implicit differentiation easy question that i can't do...

1. ## implicit differentiation easy question that i can't do...

Find the points on the graph $\displaystyle x^3 + y^3 = xy$
that are parallel to the x axis (gradient = 0)

so I have to differentiate it
$\displaystyle 3x^2 + 3y^2 (dy/dx) = dy/dx$

then what?

Thank you

2. Originally Posted by differentiate
Find the points on the graph $\displaystyle x^3 + y^3 = xy$
that are parallel to the x axis (gradient = 0)

so I have to differentiate it
$\displaystyle 3x^2 + 3y^2 (dy/dx) = dy/dx$

correction, product rule on the right hand side ...

$\displaystyle \textcolor{red}{3x^2 + 3y^2 \frac{dy}{dx} = x \frac{dy}{dx} + y}$

then what? solve for $\displaystyle \textcolor{red}{\frac{dy}{dx}}$ and set the derivative equal to 0. use that equation to find a relationship between y and x to find the points on the curve where the gradient is 0.
...

3. thank you
so then,

$\displaystyle 3x^2 + 3y^2 (dy/dx) = x.(dy/dx) + y$
$\displaystyle 3y^2(dy/dx) - x(dy/dx) = y - 3x^2$
$\displaystyle (dy/dx)(3y^2 - x) = y - 3x^2$
$\displaystyle dy/dx = (y-3x^2)/(3y^2 - x)$
dy/dx = 0
$\displaystyle y - 3x^2 = 0$
$\displaystyle y = 3x^2$

do I put that into the original equation?

$\displaystyle x^3 + (3x^2)^3 = x(3x^2)$
$\displaystyle x^3 + 27x^6 = 3x^3$
$\displaystyle 27x^6 - 2x^3 = 0$
$\displaystyle x^3(27x^3 - 2) = 0$

I'm stuck
and I don't have answers to this question

4. Originally Posted by differentiate
thank you
so then,

$\displaystyle 3x^2 + 3y^2 (dy/dx) = x.(dy/dx) + y$
$\displaystyle 3y^2(dy/dx) - x(dy/dx) = y - 3x^2$
$\displaystyle (dy/dx)(3y^2 - x) = y - 3x^2$
$\displaystyle dy/dx = (y-3x^2)/(3y^2 - x)$
dy/dx = 0
$\displaystyle y - 3x^2 = 0$
$\displaystyle y = 3x^2$

do I put that into the original equation?

$\displaystyle x^3 + (3x^2)^3 = x(3x^2)$
$\displaystyle x^3 + 27x^6 = 3x^3$
$\displaystyle 27x^6 - 2x^3 = 0$
$\displaystyle x^3(27x^3 - 2) = 0$

I'm stuck
and I don't have answers to this question
then the points are
$\displaystyle x = 0$ $\displaystyle \rightarrow y = 0$
$\displaystyle x = \frac{\sqrt[3]2}{3}$ $\displaystyle \rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2$

5. Originally Posted by dedust
then the points are
$\displaystyle x = 0$ $\displaystyle \rightarrow y = 0$
$\displaystyle x = \frac{\sqrt[3]2}{3}$ $\displaystyle \rightarrow y = \left(\frac{\sqrt[3]2}{3}\right)^2$
You mean $\displaystyle x = \frac{\sqrt[3]2}{3}$ $\displaystyle \rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2$

6. Originally Posted by pickslides
You mean $\displaystyle x = \frac{\sqrt[3]2}{3}$ $\displaystyle \rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2$
oops,..yeap