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September 1, 2014

September 1, 2014

Posted by **Ivan** on Sunday, November 28, 2010 at 10:18am.

- calculus4 -
**Reiny**, Sunday, November 28, 2010 at 10:41am6y^2 dy/dx - dy/dx = 0 - 4x^3 dx/dx

notice I wrote down dx/dx even though it has a value of 1, and i wouldn't have to include it.

Since you are differentiating with respect to x, whenever you work on a purely x-term, the dx/dx does not have to be written.

BUT, whenever you are differentiating a variable other than x, d(?)/dx has to be included.

so..

6y^2 dy/dx - dy/dx = 0 - 4x^3 dx/dx

dy/dx(6y^2 - 1) = -4x^3

dy/dx = -4x^3/(6y^2 - 1)

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