Question

If $y=\frac{(a-x)\sqrt{a-x}-(b-x)\sqrt{x-b}}{\sqrt{a-x}+\sqrt{x-b}}$, then find $\frac{dy}{dx}$ wherever defined.

• A. $\frac{2x-(a+b)}{2\sqrt{(a-x)(x-b)}}$
• B. $\frac{x}{2\sqrt{(a-x)(x-b)}}$
• C. $\frac{2x-(a+b)}{4\sqrt{(a-x)(x-b)}}$
• D. $\frac{2x+(a+b)}{2\sqrt{(a-x)(x-b)}}$

Answer 1

The correct option is A $\frac{2x-(a+b)}{2\sqrt{(a-x)(x-b)}}$

Here, $y=\frac{(a-x{)}^{3/2}+(x-b{)}^{3/2}}{(a-x{)}^{1/2}+(x-b{)}^{1/2}}$, use $a^3+b^3=(a+b)(a^2-ab+b^2)$
$y=\frac{\left\{\right(a-x{)}^{1/2}+(x-b{)}^{1/2}\}\left\{\right(a-x)-\sqrt{(a-x)(x-b)}+(x-b\left)\right\}}{\left\{\right(a-x{)}^{1/2}+(x-b{)}^{1/2}\}}$
$\Rightarrow y=(a-b)-\sqrt{(ax+bx-ab-x^2)}$
$\therefore \frac{dy}{dx}=-\frac{(a+b-2x)}{2\sqrt{(a-x)(x-b)}}\Rightarrow \frac{dy}{dx}=\frac{2x-(a+b)}{2\sqrt{(a-x)(x-b)}}$