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

**differentiate** · January 3rd 2010, 06:08 PM

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

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

then what?

Thank you

**skeeter** · January 3rd 2010, 06:20 PM

Originally Posted by differentiate

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

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

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

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

then what? solve for $\textcolor{red}{\frac{dy}{dx}}$ 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.

...

**differentiate** · January 3rd 2010, 07:42 PM

thank you
so then,

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

do I put that into the original equation?

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

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

**dedust** · January 3rd 2010, 07:56 PM

Originally Posted by differentiate

thank you
so then,

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

do I put that into the original equation?

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

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