Question

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

• A. $1$
• B. $\sqrt{\frac{a-x}{x-b}}$
• C. $\sqrt{(a-x)(x-b)}$
• D. $\frac{1}{\sqrt{(a-x)(b-x)}}$

Answer 1

The correct option is B $\sqrt{\frac{a-x}{x-b}}$

Given : $y=\sqrt{(a-x)(x-b)}-(a-b){\tan }^{-1}\sqrt{\frac{a-x}{x-b}}$
Let $x=a{\cos }^2\theta +b{\sin }^2\theta \therefore a-x=a-a{\cos }^2\theta -b{\sin }^2\theta =(a-b){\sin }^2\theta$
and
$x-b=a{\cos }^2\theta +b{\sin }^2\theta -b=(a-b){\cos }^2\theta \therefore y=(a-b)\sin \theta \cos \theta -(a-b){\tan }^{-1}\left(\tan \theta \right)=\frac{a-b}{2}\sin 2\theta -(a-b)\theta \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{(a-b)\cos 2\theta -(a-b)}{(b-a)\sin 2\theta }=\frac{1-\cos 2\theta }{\sin 2\theta }=\tan \theta =\sqrt{\frac{a-x}{x-b}}$