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July 28, 2014

July 28, 2014

Posted by **Tom** on Friday, November 11, 2011 at 11:30am.

y + y^3 + 3 = e^y^2 + 3^x * cos(3y) - x

Evaluate dy/dx

I get to:

dy/dx (1+3y) = e^y^2 * 3y^2 * dy/dx + 3^x * ln3 * (-sin(3y)) * 3 *dy/dx -1

Will it give me?

dy/dx(1+3y) =

dy/dx((e^y^2 * 3y^2) +

(3^x * ln3 * (-sin(3y)) * 3 ) -1

dy/dx(1+3y) =

dy/dx((e^y^2 * 3y^2) - (9^xsin(3y)ln3))

-1

1 / ((e^y^2 * 3y^2) - (9^xsin(3y)ln3) - (1+3y))

- calculus -
**Tom**, Friday, November 11, 2011 at 11:31amlast one is supposed to be

dy/dx = 1 / ((e^y^2 * 3y^2) - (9^xsin(3y)ln3) - (1+3y))

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