$\displaystyle x^3-6y^2=10$ $\displaystyle 3x^2-(12y)\frac{dy}{dx}=0$ $\displaystyle 3x^2=(12y)\frac{dy}{dx}$ $\displaystyle \frac{3x^2}{12y}=\frac{dy}{dx}$
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Just to simplify it further, $\displaystyle \frac{x^{2}}{4y}$, otherwise, absolutely correct.
Thanks man for all the help with these!
No problem.
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