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April 18, 2014

April 18, 2014

Posted by **marcus** on Monday, December 8, 2008 at 4:02am.

- calculus -
**drwls**, Monday, December 8, 2008 at 6:19amUse implicit differentiation.

Differentiate both sides of the equation with respect to x, treating y as a function of x.

x cosy dy/dx + sin y - 2 sin 2y dy/dx = -sin y dy/dx

Now solve for dy/dx. It will be a function of both x and y.

You could also differentiate both sides with respect to y, solve for dx/dy, and take the reciprocal of the answer. The result will look different but will still be correct.

x cos y + sin y dx/dy - 2 sin 2y = -sin y

dx/dy = [-siny + 2 sin2y -x cos y]/sin y

dy/dx = sin y/[-siny + 2 sin2y -x cos y]

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