Question

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

Answer 1

Let $f\left(x\right)=\frac{\sin x+\cos x}{\sin x-\cos x}$
Differentiating with respect to $x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{\sin x+\cos x}{\sin x-\cos x}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(\sin x-\cos x)\frac{d}{dx}(\sin x+\cos x)-(\sin x+\cos x)\frac{d}{dx}(\sin x-\cos x)}{(\sin x-\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(\cos x-\sin x)(\sin x-\cos x)-(\cos x+\sin x)(\sin x+\cos x)}{(\sin x-\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-(\sin x-\cos x)(\sin x-\cos x)-(\sin x+\cos x)(\sin x+\cos x)}{(\sin x-\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-(\sin x-\cos x{)}^2-(\sin x+\cos x{)}^2}{(\sin x-\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-\left[\right({\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x)+({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x\left)\right]}{(\sin x-\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-(2{\sin }^{2}x+2{\cos }^{2}x+0)}{(\sin x-\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-2({\sin }^{2}x+{\cos }^{2}x)}{(\sin x-\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-2\left(1\right)}{(\sin x-\cos x{)}^2}$
$\therefore f^{\prime }\left(x\right)=\frac{-2}{(\sin x-\cos x{)}^2}$