Find d y d x if y= ^ -1 ( 3-x 3+x ).

Question

Find $\frac{d y}{d x}$ if $y=\tan ^{-1}\left(\sqrt{\frac{3-x}{3+x}}\right)$.

✓

Answer

$y=\tan ^{-1}\left(\sqrt{\frac{3-x}{3+x}}\right)$
Put $x=3 \cos 2 \theta$
$\therefore \quad \theta=\frac{1}{2} \cos ^{-1}\left(\frac{x}{3}\right)\quad\quad\ldots\ldots (1)$
$y=\tan ^{-1}\left(\sqrt{\frac{3-3 \cos 2 \theta}{3+3 \cos 2 \theta}}\right)$
$=\tan ^{-1}\left(\sqrt{\frac{3(1-\cos 2 \theta)}{3(1+\cos 2 \theta)}}\right)$
$=\tan ^{-1}\left(\sqrt{\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}}\right) \quad \ldots \ldots \ldots\left[\begin{array}{ll}\because & 1-\cos 2 \theta=2 \sin ^2 \theta \text { and } 1+\cos 2 \theta=2 \cos ^2 \theta\end{array}\right]$
$\therefore \quad=\tan ^{-1}\left(\sqrt{\tan ^2 \theta}\right)$
$\therefore \quad=\tan ^{-1}(\tan \theta)$
$\therefore \quad y=\theta \quad \ldots\ldots \left[\because \quad \tan ^{-1}(\tan \theta)=\theta\right]$
$\therefore \quad y=\frac{1}{2} \cos ^{-1}\left(\frac{x}{3}\right) \quad\ldots\ldots$ [From(1)]
Differentiate w.r.t. $x$
$\frac{d y}{d x}=-\frac{1}{2} \cdot \frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^2}} \times \frac{1}{3} \quad \cdots \cdots \cdot\left[\because \quad \frac{d}{d x}\left(\cos ^{-1} x\right)=-\frac{1}{\sqrt{1-x^2}}\right]$
$\therefore \quad \frac{d y}{d x}=-\frac{1}{6} \cdot \frac{1}{\sqrt{9-x^2}} \times 3$
$\therefore \quad \frac{d y}{d x}=-\frac{1}{2 \sqrt{9-x^2}}$