Posted by **lil** on Wednesday, September 27, 2006 at 3:54pm.

what is the derivative of sinx- cosy = 0 in its simplest form? I got to the point y'=-consx/siny, but I wasn't too sure. This could probably simplified further. help!

You're supposed to treat y as a function of x and differentiate it implicitly. We have

d/dx{sinx- cosy)=d/dx{sinx) - d/dx{cosy)=

cos(x)-(-sin(y)*y')=0

You should be able to solve for y'.

I did, and I ended up with y'=-cos(x)/sin(y)

How do I simplify this further?

That's as far as you can go unless you know something specific about y. Without knowing something about y(x) there's nothing else to do here.

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