I've worked through this problem a few times, haven't gotten the correct answer my book has, and am thinking I might have an issue in my process. Would someone be willing to take a look at my work, and show me what I'm doing wrong?
The problem: 4cos(x)sin(y) = 1; Find dy/dx with Implicit Differentiation.

Any and all help is greatly appreciated!

Question

I've worked through this problem a few times, haven't gotten the correct answer my book has, and am thinking I might have an issue in my process. Would someone be willing to take a look at my work, and show me what I'm doing wrong? 
The problem: 4cos(x)sin(y) = 1; Find dy/dx with Implicit Differentiation.

Any and all help is greatly appreciated!

amonoconnor · Answer

So initially, I am thinking I need to use the Product Rule, in combination with the Chain Rule, making it f(x)g(x), with f(x)=4cos(x) and g(x)=sinx(y). I used the Product Rule on f(x), plugging in f'(x) as a whole for f'(x) when I used the product rule on f(x)g(x).

aryavegapunk · Answer

4 * cos(x) * sin(y) = 1 
cos(x) * sin(y) = 1/4 

Just to make it a bit easier

anonymous · Answer

Try
$$4\sin y=\sec x$$ and follow the rules for implicit dif.

aryavegapunk · Answer

Next, would be the product rule,

cos(x) * cos(y) * dy - sin(y) * sin(x) * dx = 0 
dy * cos(x) * cos(y) = dx * sin(y) * sin(x)

aryavegapunk · Answer

dy/dx = (sin(y) * sin(x)) / (cos(y) * cos(x))

aryavegapunk · Answer

Did that help? There's one last step, but I was going to leave that to you

amonoconnor · Answer

I'll erase my work, and try it again:) Give me just one sec! :)

amonoconnor · Answer

I get, based on this method, dy/dx=tan(x)tan(y)... is that right?

aryavegapunk · Answer

yes that is correct

amonoconnor · Answer

Would you be willing to give me a pointer or two on another problem, on the same topic? I'll post it in as a new question.

aryavegapunk · Answer

I'll try. Forgive me if I can't though. I actually did this same problem when I took Calc 1