Find the derivative of the function: $\frac{\sin x+\cos x}{\sin x-\cos x}$

Let $f(x) = \frac{\sin x+\cos x}{\sin x-\cos x}$.
Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:
$f'(x) = \frac{(\sin x - \cos x) \frac{d}{dx}(\sin x + \cos x) - (\sin x + \cos x) \frac{d}{dx}(\sin x - \cos x)}{(\sin x - \cos x)^2}$
$f'(x) = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}$
$f'(x) = \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2}$
$f'(x) = \frac{-(\sin^2 x + \cos^2 x - 2\sin x \cos x) - (\sin^2 x + \cos^2 x + 2\sin x \cos x)}{(\sin x - \cos x)^2}$
Since $\sin^2 x + \cos^2 x = 1$:
$f'(x) = \frac{-(1 - 2\sin x \cos x) - (1 + 2\sin x \cos x)}{(\sin x - \cos x)^2}$
$f'(x) = \frac{-1 + 2\sin x \cos x - 1 - 2\sin x \cos x}{(\sin x - \cos x)^2}$
$f'(x) = \frac{-2}{(\sin x - \cos x)^2}$