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  4. If y=√(a-x)(x-b)-(a-b) tan -1 √(a-x/x-b), then (d y/d x) is equal to

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Q. If $y=\sqrt{(a-x)(x-b)}-(a-b) \tan ^{-1} \sqrt{\frac{a-x}{x-b}}$, then $\frac{d y}{d x}$ is equal to

ManipalManipal 2013

Solution:

Let $x=a\, \cos ^{2} \theta+b \,\sin ^{2} \theta$
$\therefore a-x=a-a \cos \theta-b \sin ^{2} \theta=(a-b) \sin ^{2} \theta$
and $x-b=a \cos ^{2} \theta-b \sin ^{2} \theta-b=(a-b) \cos ^{2} \theta$
$\therefore y=(a-b) \sin \theta \cdot \cos \theta-(a-b) \tan ^{-1} \tan \theta$
$=\frac{a-b}{2} \sin 2 \theta-(a-b) \theta$
$\therefore \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{(a-b) \cos 2 \theta-(a-b)}{(b-a) \sin 2 \theta}$
$=\frac{1-\cos 2 \theta}{\sin \theta}=\tan \theta=\sqrt{\frac{a-x}{x-b}}$