$\displaystyle x^3-6y^2=10$

$\displaystyle 3x^2-(12y)\frac{dy}{dx}=0$

$\displaystyle 3x^2=(12y)\frac{dy}{dx}$

$\displaystyle \frac{3x^2}{12y}=\frac{dy}{dx}$

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- Mar 22nd 2009, 07:09 PMJim MarnellLast dy/dx Question. Just needs checked.
$\displaystyle x^3-6y^2=10$

$\displaystyle 3x^2-(12y)\frac{dy}{dx}=0$

$\displaystyle 3x^2=(12y)\frac{dy}{dx}$

$\displaystyle \frac{3x^2}{12y}=\frac{dy}{dx}$ - Mar 22nd 2009, 07:13 PMPinkk
Just to simplify it further, $\displaystyle \frac{x^{2}}{4y}$, otherwise, absolutely correct.

- Mar 22nd 2009, 07:17 PMJim Marnell
Thanks man for all the help with these!

- Mar 22nd 2009, 07:21 PMPinkk
No problem. (Cool)