Question

If $y=\frac{\sin x+\cos x}{\sin x-\cos x}$ find $\frac{dy}{dx}$.

Answer 1

$y=\frac{\sin x+\cos x}{\sin x-\cos x}=\frac{\tan x+1}{\tan x-1}$

$\frac{dy}{dx}=\frac{(\tan x-1)\left({\sec }^{2}x\right)-(\tan x+1)\left({\sec }^{2}x\right)}{{(\tan x-1)}^2}$

$=\frac{2{\sec }^{2}x}{{(\tan x-1)}^2}=\frac{2{\sec }^{2}x}{{\sec }^{2}x-2\tan x}$

$=\frac{\frac{2}{{\cos }^{2}x}}{\frac{1}{{\cos }^{2}x}-\frac{2\sin x}{\cos x}}$

$=\frac{2}{1-2\sin x\cos x}=\frac{2}{1-\sin 2x}$