x^3 * y^3 - y = x help please differentiate implicitly with respect to x...
Last edited by Jhevon; Mar 9th 2009 at 05:09 PM.
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Originally Posted by edge x^3 * y^3 - y = x help please $\displaystyle x^3 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 3x^2 - \frac{dy}{dx}=1$ remember to use product rule on the first.
Okay, thanks... now just to make sure I'm doing this right, how would you solve for dy/dx?
Originally Posted by GaloisTheory1 $\displaystyle x^3 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 3x^2 - \frac{dy}{dx}=1$ remember to use product rule on the first. $\displaystyle x^3 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 3x^2 - \frac{dy}{dx}=1$ so $\displaystyle \frac{dy}{dx}=\frac{1-y^3 \cdot 3x^2}{x^3 \cdot 3y^2-1}$
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