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April 16, 2014

April 16, 2014

Posted by **C** on Tuesday, March 26, 2013 at 8:55pm.

- Calculus -
**Reiny**, Tuesday, March 26, 2013 at 9:24pmTwo assumptions

1. you want dy/dx

2. You have the equation 5x^2 - 2xy + 7y^2 = 0

10x - 2x dy/dx - 2y + 14y dy/dx = 0

dy/dx( 14y - 2x) = 2y - 10x

dy/dx = (2y-10x)/(14y - 2x) = (y - 5x)/(7y - x)

- Calculus -
**C**, Tuesday, March 26, 2013 at 9:27pmNo, the equation is

5x^2-2xy+7^y

Last term has 7 to the y power....

- Calculus -
**Reiny**, Tuesday, March 26, 2013 at 9:43pmOuch, nasty term ...

the derivative of 7^y would be

(ln7)(7^y)dy/dx

I am sure you can make the necessary changes .

- Calculus -
**C**, Tuesday, March 26, 2013 at 9:45pmYes...very nasty! Thanks

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