Question
Diffrentiate the following w.r.t.x
$\log \left(\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}}\right)$
$\log \left(\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}}\right)$
Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\log \tan \left(\frac{3 x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{3 x}{2}\right)} \times \frac{d}{d x}\left[\tan \left(\frac{3 x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{3 x}{2}\right)} \times \sec ^2\left(\frac{3 x}{2}\right) \cdot \frac{d}{d x}\left(\frac{3 x}{2}\right)\end{aligned}$
$\begin{aligned} & =\frac{\cos \left(\frac{3 x}{2}\right)}{\sin \left(\frac{3 x}{2}\right)} \times \frac{1}{\cos ^2\left(\frac{3 x}{2}\right)} \times \frac{3}{2} \times 1 \\ & =3 \times \frac{1}{2 \sin \left(\frac{3 x}{2}\right) \cos \left(\frac{3 x}{2}\right)} \\ & =3 \times \frac{1}{\sin 3 x}=3 \operatorname{cosec} 3 x .\end{aligned}$
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$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$