If $\displaystyle y = \frac{\sin x}{1 + \frac{\cos x}{1 + \frac{\sin x}{1 + \frac{\cos x}{1 + \sin x\mbox{...}\infty}}}}$, then prove that $\displaystyle \frac{dy}{dx} = \frac{(1 + y)\cos x + y\sin x}{1 + 2y + \cos x - \sin x}$
$\displaystyle y=\frac{sin x}{1+\frac{cos x}{1+y}}$
$\displaystyle y=\frac{sin x(1+y)}{1+y+cos x}$
$\displaystyle y+y^2+ycos x=sin x+ysin x$
$\displaystyle \frac{dy}{dx}+2y\frac{dy}{dx}+y*-sin x+cos x*\frac{dy}{dx}=cos x+ycos x+sin x\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx}(1+2y+cos x)-ysin x=cos x(1+y)+sin x\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx}(1+2y+cos x-sin x)=(1+y)cos x+y sin x$
$\displaystyle \frac{dy}{dx}(1+2y+cos x-sin x)=(1+y)cos x+y sin x$
$\displaystyle \frac{dy}{dx}=\frac{(1+y)cos x+y sin x}{1+2y+cos x-sin x}$
To write continued fractions one can use \cfrac instead of \frac, it makes the fraction more readable.
$\displaystyle y = \cfrac{\sin x}{1 + \cfrac{\cos x}{1 + \cfrac{\sin x}{1 + \cfrac{\cos x}{1 + \cfrac{\sin x}{1 + \cfrac{\cos x}{1 + \cfrac{\sin x}{1 + \cfrac{\cos x}{1 + \ldots}}}}}}}}$