Question

Find the derivative of the following (it is to be understood that $a,b,c,d,p,q,r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $\frac{\sin x+\cos x}{\sin x-\cos x}$

Answer 1

Let $f\left(x\right)=\frac{\sin x+\cos x}{\sin x-\cos x}$
Thus by quotient rule,
$f^{\prime }\left(x\right)=\frac{(\sin x-\cos x)\frac{d}{dx}(\sin x+\cos x)-(\sin x+\cos x)\frac{d}{dx}(\sin x-\cos x)}{{(\sin x-\cos x)}^2}$
$=\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x-\cos x)(\cos x-\sin x)}{{(\sin x-\cos x)}^2}$
$=\frac{-{(\sin x-\cos x)}^2-(\sin x+\cos x)}{{(\sin x-\cos x)}^2}$
$=\frac{-[{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x+{\sin }^2+{\cos }^{2}x+2\sin x+\cos x]}{{(\sin x-\cos x)}^2}$
$=\frac{-[1+1]}{{(\sin x-\cos x)}^2}=\frac{-2}{{(\sin x-\cos x)}^2}$