During implicit differentiation of this problem: $y^2 = x^2 + \sin(xy)$
I cannot figure out why at this point in differentiation this:
$$2y\frac{dy}{dx}= 2x + \left(\cos xy\right) \left(y+x\frac{dy}{dx}\right)$$
changes to:
$$2y\frac{dy}{dx}- \left(\cos xy\right) \left(x\right)\frac{dy}{dx}= 2x + \left(\cos xy\right)y$$
If they subtracted the $\cos xy$ from one side why is it still on the other?
\begin{align*} 2y\frac{\mathrm dy}{\mathrm dx} &= 2x + (\cos xy)(y+x\frac{\mathrm dy}{\mathrm dx}) &&\text{Initial equation} \\ 2y\frac{\mathrm dy}{\mathrm dx} &= 2x + (\cos xy)(y) + (\cos xy)(x)\frac{\mathrm dy}{\mathrm dx} &&\text{Expand \((y+x\frac{\mathrm dy}{\mathrm dx})\)} \\ 2y\frac{\mathrm dy}{\mathrm dx} - (\cos xy)(x)\frac{\mathrm dy}{\mathrm dx} &= 2x + (\cos xy)(y) &&\text{Subtract \((\cos xy)(x)\frac{\mathrm dy}{\mathrm dx}\)} \end{align*}
The point to note: $\cos xy$ is not standing on its own; instead, it is multiplied by a polynomial which, when expanded, becomes the sum of two terms:
$$ (\cos xy)(y) + (\cos xy)(x)\frac{\mathrm dy}{\mathrm dx} $$
both of which having $\cos xy$ as a factor. Thus, the term that is subtracted from both sides (and eliminated on the right side) of the equation is not $\cos xy$ but instead $(\cos xy)(x)\frac{\mathrm dy}{\mathrm dx}$.