Question

If $y=\log \sqrt{\frac{1-\cos 3x}{1+\cos 3x}}$, find $\frac{dy}{dx}$.

Answer 1

If $y=\log \sqrt{\frac{1-\cos 3x}{1+\cos 3x}}$

$y=\log {\left(\frac{1-\cos 3x}{1+\cos 3x}\right)}^{1/2}$

$y=\frac{1}{2}\left[\log \right(1-\cos 3x)-\log (1+\cos 3x\left)\right]$

Differentiate w.r. to $x$ both sides

$\frac{dy}{dx}=\frac{1}{2}\left[\frac{1}{1-\cos 3x}\left(3\sin 3x\right)-\frac{-3\sin 3x}{1+\cos 3x}\right]$

$\frac{dy}{dx}=\frac{3\sin 3x}{2}\left[\frac{1}{1-\cos 3x}+\frac{1}{1+\cos 3x}\right]$

$\frac{dy}{dx}=\frac{3\sin 3x}{2}\left[\frac{1+\cos 3x+1-\cos 3x}{(1-\cos 3x)(1+\cos 3x)}\right]$

$\frac{dy}{dx}=\frac{3\sin 3x}{2}\times \frac{2}{1-{\cos }^{2}3x}$

$\frac{dy}{dx}=\frac{3\sin 3x}{1-{\cos }^{2}3x}$

$\frac{dy}{dx}=\frac{3\sin 3x}{{\sin }^{2}3x}=\frac{3}{\sin 3x}=3\text{cosec}3x$.