Implicit differentiation $(\sin x +y)/(\cos y +x)=1$ I am currently stuck on this problem

Find $\frac{dy}{dx}$ when
$$\frac{\sin x+y}{\cos y+x}=1$$

I've been using the quotient rule to try and differentiate this, but my online homework rejected my answer of:
$$\frac{-x\cos x - \cos x\cos y +y+\sin x}{\cos y +x-\sin y \sin x+y\sin y}$$
Can anyone tell me where I went wrong, or if there is another way to do this besides the quotient rule?
 A: Actually, Quotient Rule makes it hell complicated. The simple and intended way would be writing the given equation as
$$\sin x+y=\cos y +x$$
And then differentiate both sides w.r.t. $x$ to get
$$\cos x+\frac{dy}{dx}=-\sin y\frac{dy}{dx}+1$$
$$\implies (1+\sin y)\frac{dy}{dx}=1-\cos x$$
$$\implies \frac{dy}{dx}=\frac{1-\cos x}{1+\sin y}$$
which should be accepted as the answer.
Another thing, if you did it the correct way, then after some damn long simplifications, you will arrive at the same result using Quotient Rule.
Hope this helps :)
A: The use of the Quotient Rule doesn't need to be complicated, as long as one resists differentiating any expressions that don't really need that (or writing out components until we finally need them).  If we call the ratio $ \ \large{\frac{\sin x \ + \ y}{\cos y \ + \ x}} \ \normalsize{= \ \frac{f(x)}{g(x)}} \ = \  1 \ \ , $ then we can write
$$ \frac{g(x) · (\cos x \ + \ y') \ - \ f(x) · (-\sin y · y' \ + \ 1)}{[ \ g(x) \ ]^2} \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \frac{g(x) ·   y'  \ + \ f(x) ·  \sin y · y' }{[ \ g(x) \ ]^2} \ \ = \ \ \frac{-g(x) ·  \cos x   \ + \ f(x) }{[ \ g(x) \ ]^2}  $$
$$ \Rightarrow \ \ y' \ \ = \ \ \frac{f(x) \ - \ g(x) ·  \cos x   }{g(x)   \ + \ f(x) ·  \sin y} \ \ = \ \ \frac{ (\sin x \ + \ y) \ - \  (\cos y \ + \ x) ·  \cos x   }{ (\cos y \ + \ x)   \ + \  (\sin x \ + \ y) ·  \sin y} $$
$$ = \ \ \frac{  \sin x \ + \ y  \ - \  \cos x \cos y \ - \ x  \cos x   }{ \cos y \ + \ x   \ + \  \sin x \sin y \ + \ y \sin y} \ \ , $$
so you were nearly correct.  Raffaele's comment about the sign error in the denominator is confirmed and this may well be all that the on-line system was "rejecting."
To "bridge" this to ultralegend5385's much simpler expression for the derivative, we will want to use the original equation, which tells us that $ \ \sin x \ + \ y \ = \ \cos y \ + \ x \ \ . $  Upon checking the above result for the appearance of such sums, we observe that
$$ y' \ \ = \ \ \frac{  ( \ \sin x \ + \ y \ )  \ - \  [ \ \cos x \ · \ ( \ \cos y \ + \ x \ ) \ ]  }{ ( \ \cos y \ + \ x \ )   \ + \  [ \ ( \ \sin x   \ + \ y \ ) · \ \sin y \ ]} \ \ = \ \ \frac{  ( \ \sin x \ + \ y \ )  \ - \  [ \ \cos x \ · \ ( \ \sin x \ + \ y \ ) \ ]  }{ ( \ \sin x \ + \ y \ )   \ + \  [ \ ( \ \sin x   \ + \ y \ ) · \ \sin y \ ]}  $$
$$ = \ \ \frac{  ( \ \sin x \ + \ y \ ) \ · \ ( \ 1  \ - \    \cos x \ )  }{ ( \ \sin x \ + \ y \ ) \ · \ ( \ 1  \ + \    \ \sin y  \ ) } \ \ .  $$
