Question

If $y=\sqrt{(a-x)(x-b)}-(a-b)ta{n}^{-1}\sqrt{\frac{a-x}{a-b}},then\frac{dy}{dx}=$

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

Answer 1

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

$y=\sqrt{(a-x)(x-b)}-(a-b)t{a}^{-1}\sqrt{\frac{a-x}{x-b}}$
$\frac{d_y}{d_x}=\frac{1}{2\sqrt{(a-x)(a-b)}}\times \frac{d}{d_n}\left[\right(a-x\left)\right(a-b\left)\right]-(a-b)\frac{1}{1+\frac{a-x}{x-b}}\times \frac{d}{d_n}\sqrt{\frac{a-2}{x-b}}$
$=\frac{1}{2\sqrt{(a-x)(x-b)}}\times [-1(x-b)+1(a-x\left)\right]-(a-b)\frac{1}{\frac{x-b+a-x}{x-b}}\left[\frac{1}{2\sqrt{\frac{a-x}{x-b}}\times \frac{d}{d_n}\frac{a-x}{x-b}}\right]=\frac{1}{2\sqrt{(a-x)(x-b)}}[-x+b+a-x]-(a-b)\frac{x-b}{a-b}[\frac{1}{2}\sqrt{\frac{x-b}{a-x}}\times \frac{-1(x-b)-(a-x)}{(x-b{)}^2}]=\frac{a+b-2x}{2\sqrt{(a-2)(x-b)}}-\frac{1}{2}\sqrt{\frac{x-b}{a-x}}\times \frac{b-a}{(x-b)}$
$=\frac{a+b-2x}{2\sqrt{(a-x)(x-b)}}-\frac{1}{2}\frac{b-a}{\sqrt{a-x}\sqrt{x-b}}=\frac{a+b-2x-b+a}{2\sqrt{(a-x)(x-b)}}=\frac{2a-2x}{2\sqrt{(a-x)(x-b)}}=\frac{2(a-x)}{2\sqrt{(a-x)(x-b)}}=\frac{\sqrt{a-x}}{\sqrt{x-b}}=\sqrt{\frac{a-x}{}x-b}$