Question

Find the derivative of following functions w.r.t. $x$:
${\log }_c\sqrt{\frac{1-\cos x}{1+\cos x}}$

Answer 1

Let $y={\log }_c\sqrt{\frac{1-\cos x}{1+\cos x}}$

$=\left[\frac{1}{2}{\log }_e\left(\frac{1-\cos x}{1+\cos x}\right)\right]$$[\because \log \sqrt{t}=\frac{1}{2}\log t]$

Diff. w.r.t. $x$,

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

$=\frac{1}{2}\left[\frac{1}{1-cosx}\frac{d}{dx}\right(1-\cos x)-\frac{1}{1+cosx}\frac{d}{dx}(1+\cos x\left)\right]$

$=\frac{1}{2}\left[\frac{1}{1-cosx}\right(\sin x)-\frac{1}{1+cosx}(-\sin x\left)\right]$

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

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

$=\frac{1}{2}\frac{\sin x\left[2\right]}{{\sin }^{2}x}=\frac{2\sin x}{2{\sin }^{2}x}=\csc x$