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October 30, 2014

October 30, 2014

Posted by **Cupcake** on Tuesday, April 1, 2008 at 1:26pm.

I would appreciate any help that can be offered.

- Calculus 1 - implicit differentiation -
**drwls**, Tuesday, April 1, 2008 at 1:38pmJust differtiate each term with respect to x, remembering that y is a function of x. You often end up with an expression for y' that involves both x and y, but is is a valid equation. You may have to evaluate y before you can calculate y'. Here is what you get:

x*dy/dx + y + 2 + 6x = 0

dy/dx = -(y + 2 + 6x)/x

= -(y+2)/x -6

- Calculus 1 - implicit differentiation -
**Cupcake**, Tuesday, April 1, 2008 at 2:28pmthank you

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