Question

If $y=\sqrt{(a-x)(x-b)}-(a-b){\tan }^{-1}\sqrt{\frac{a-x}{x-b}}(a>b)$ then $\frac{dy}{dx}=$.

• A. $\sqrt{\frac{a-x}{x-b}}$
• B. $\sqrt{(a-x)(x-b)}$
• C. $0$
• D. $1$

Answer 1

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

Given, $y=\sqrt{(a-x)(x-b)}-(a-b){\tan }^{-1}\sqrt{\frac{a-x}{x-b}}$Let $u=(a-x)(x-b),\surd =\sqrt{\frac{a-x}{x-b}}$$\Rightarrow y=\sqrt{u}-(a-b){\tan }^{-1}v$Now, differentiating wrt $x$$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\sqrt{u}\right)-(a-b)\frac{d}{dx}{\tan }^{-1}v$$=\frac{d\left(\sqrt{u}\right)}{du}.\frac{du}{dx}-(a-b)\frac{d{\tan }^{-1}v}{dv}.\frac{dv}{dx}$ [chain rule]$=\frac{1}{2\sqrt{u}}.\frac{du}{dx}-(a-b)\frac{1}{1+v^2}.\frac{dv}{dx}......\left(1\right)$$\frac{du}{dx}=\frac{d}{dx}(a-x)(x-b)=\frac{d}{dx}(ax-av-x^2+bx)=a-2x+b$$\frac{dv}{dx}=\frac{d}{dx}\left(\sqrt{\frac{a-x}{x-b}}\right)=\frac{1}{2\sqrt{\frac{a-x}{x-b}}}\times \frac{(x-b)(-1)-(a-x)\left(1\right)}{(x-b{)}^2}$ [quotent rule]$=\frac{\sqrt{x-b}}{2\sqrt{a-x}}\times \frac{(b-a)}{(x-b{)}^2}$substituting in $\left(1\right)$$\frac{dy}{dx}=\frac{1}{2\sqrt{(a-x)(x-b)}}\times (a-2x+b)-(a-b)\frac{1}{1+\frac{a-x}{x-b}}\times \frac{\sqrt{x-b}}{\sqrt{b-x}}\times \frac{(b-a)}{(x-b{)}^2}$$=\frac{a-2x+b}{2\sqrt{(a-x)(x-b)}}-\frac{\left(ab\right)\left(xb\right)}{(a-b)}\times \frac{\sqrt{x-b}}{2\sqrt{a-x}}\frac{(b-a)}{(x-b{)}^2}$$=\frac{a-2x+b+a-b}{2\sqrt{(a-x)(x-b)}}$$=\frac{2(a-x)}{2\sqrt{(a-x)(x-b)}}$$\Rightarrow \frac{dy}{dx}=\sqrt{\frac{a-x}{x-b}}$option $A$ is corect