Some implicit differentiation questions. Please check them [see desc.] I have a few implicit diffentiatial questions that I wanted to check. general question... how do you know that $y$ is a function of $x$? I assume the whole reason why these are called implicit differentiation questions is because $y$ is defined implicitly and it's hard or impossible to define $y$ as a function of $x$. Is that right?
1.
$$ \cos{xy} = 1+ \sin{y} $$
$$-\sin{xy} \cdot (\frac{dy}{dx} + y) = \cos{y} \cdot \frac{dy}{dx}$$
$$-\sin{xy} \cdot \frac{dy}{dx} - y(\sin{xy}) = \cos{y} \cdot \frac{dy}{dx}$$
$$ - y(\sin{xy}) = \cos{y} \cdot \frac{dy}{dx} + \sin{xy} \cdot \frac{dy}{dx}$$
$$ - y(\sin{xy}) = \frac{dy}{dx} ( \sin{xy} + \cos{y})$$
$$ \frac{- y(\sin{xy})}{( \sin{xy} + \cos{y})} = \frac{dy}{dx}$$


*$$x - y = x \cdot e^y$$
$$1 - \frac{dy}{dx} = x \cdot e^y \cdot \frac{dy}{dx} + e^y$$
$$ 1 - e^y = x \cdot e^y \frac{dy}{dx} + \frac{dy}{dx}$$
$$1 - e^y = \frac{dy}{dx}(x \cdot e^y + 1)$$
$$\frac{1-e^y}{x \cdot e^y + 1} = \frac{dy}{dx}$$

*$$y \cdot \cos{x} = x^2 + y^2$$
$$y(-\sin{x}) \cdot \cos{x} \cdot \frac{dy}{dx} = 2x + 2y \frac{dy}{dx}$$
$$ 2y \frac{dy}{dx} - \cos{x} \frac{dy}{dx} = y - \sin{x} - 2x$$
$$\frac{dy}{dx} (2y-cosx) = y - sinx - 2x$$
$$\frac{dy}{dx} = \frac{y - sinx - 2x}{2y - cosx}$$
Thank you.
 A: 1) As John pointed there is a mistake...
2) The second example seems correct to me
3) For the third differenciation...
$$y \cdot \cos{x} = x^2 + y^2$$
$$y'\cos(x)-y\sin(x)=2x+2yy'$$
$$y'\cos(x)-2yy'=y\sin(x)+2x$$
$$y'(\cos(x)-2y)=y\sin(x)+2x$$
$$\frac {dy}{dx}=\frac {y\sin(x)+2x}{\cos(x)-2y} $$
...
A: Note that in the case of implicit differentiation it is not necessarily the case that one variable is a function of the other in the strict sense, but the two functions are related. One can symbolize this as $x$ and $y$ being parametric functions of a variable $t$. An alternate approach to implicit differentiation is based on this idea.
Taking your first example:
\begin{eqnarray}
\cos{xy} &=& 1+ \sin{y}\\
\frac{d}{dt}\left(\cos{xy} \right)&=&\frac{d}{dt}\left(1+ \sin{y}\right)\\
-\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right)\sin xy&=&(\cos y)\frac{dy}{dt}\\
-(x\,dy+y\,dx)\sin xy&=&\cos y\,dy\\
-y\sin xy\,dx&=&(x\sin xy+\cos y)\,dy\\
\frac{dy}{dx}&=&-\frac{y\sin xy}{x\sin xy+\cos y}
\end{eqnarray}
