Question

If $y=log\sqrt{\frac{a\cos x-b\sin x}{a\cos x+b\sin x}}$, prove that $\frac{dy}{dx}=\frac{-ab}{a^2{\cos }^{2}x-b^2{\sin }^{2}x}$.

Answer 1

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

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

$\frac{dy}{dx}=\frac{1}{2}[\frac{1(-a\sin x-b\cos x)}{a\cos x-b\sin x}-\frac{1.(-a\sin x+b\cos x)}{(a\cos x+b\sin x)}]$

$\frac{dy}{dx}=\frac{1}{2}[-\frac{(a\sin x+b\cos x)(a\cos x+b\sin x)-(a\cos x-b\sin x)(b\cos x-a\sin x)}{a^{2}co{s}^{2}x-b^2{\sin }^{2}x}]$

$\Rightarrow \frac{1}{2}[-\frac{2ab{\sin }^2\theta -2ab{\cos }^{2}x}{a^2{\cos }^{2}x-b^2{\sin }^{2}x}]=\frac{1}{2}[-2ab\frac{(si{n}^{2}x+{\cos }^{2}x)}{a^2{\cos }^{2}x-b^2{\sin }^{2}x}]]$

$\therefore \frac{dy}{dx}=\frac{-2ab}{2(a^2{\cos }^{2}x-b^2{\sin }^{2}x)}=\frac{-ab}{a^2{\cos }^{2}x-b^2{\sin }^{2}x}$