Question

Differentiate w.r.t. x: $\left(\frac{sinx+cosx}{sinx-cosx}\right)$

Answer 1

We have,

$\frac{d}{dx}\left(\frac{sinx+cosx}{sinx-cosx}\right)$

$=\frac{\left(\cos x-\sin x\right)\left(sinx-cosx\right)-\left(\sin x+\cos x\right)\left(\cos x+\sin x\right)}{{\left(\sin x-\cos x\right)}^2}$

$=\frac{\left(\sin x\cos x-{\cos }^{2}x-{\sin }^{2}x+\sin x\left(\cos x\right)-{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x\right)}{{\left(\sin x-\cos x\right)}^2}$

$=\frac{-2}{{\left(\sin x-\cos x\right)}^2}$