implicit differentiation of 9 cosx siny = 8

I

integragirl

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Apr 13, 2006
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please help, 9 cosx siny=8
I know that to begin you take the derivative of both sides but I'm not sure where to go next. Do I use the product rule. Any help would be appreciated
 
Yes use the product rule.

9(cosx)(cosy)+9(siny)(-sinx)=0

I hope you don't have to solve y, because at the moment it looks too complicated for me...
 
Is there a product?

Hint: (d/dy)sin(y) = cos(y)(dy/dx)

Posting the same question, after you already received a useful answer, really is not good form.
 
I didn't say that I didn't post it on other completely different websites. I do this not in bad taste but to get help (as in different peoples opinion) and to get the fastest response possible. I see nothing wrong with that.
 
integragirl said:
I didn't say that I didn't post it on other completely different websites. I do this not in bad taste but to get help (as in different peoples opinion) and to get the fastest response possible. I see nothing wrong with that.
Click to expand...

Sure there is. It's a waste of people's time and very selfish. :roll:
 
You will also note that I did NOT say you had already posted HERE, only that you had posted.

I understand your motivations. They are very common. It is time for you to see something wrong with it. Plenty of folks visit many sites. If you want additional opinions, you can ask. If you want more clarification, you can ask. If you demand instantaneous response, you should seek PAID answers. Volunteers just don't work that way.

Just learn. We can all get along just fine if we learn.

Just for the record, you do NOT need the product rule. You can use it if you wish. An alternate method is about as complicated.

Rewrite to: \(\displaystyle \L\,sin(y)\,=\,\frac{8}{9}*\frac{1}{cos(x)}\)

Then you've only a chain rule and a little algebra.
 
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