If y = frac{(a - x)sqrt{a - x} - (b - x)sqrt{ | vedclass

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(B) Let $u = \sqrt{a - x}$ and $v = \sqrt{x - b}$. Then $u^2 = a - x$ and $v^2 = x - b$.Note that $u^2 + v^2 = a - b$. The numerator is $u^3 - (b - x)

Vedclass · Accepted Answer

$\frac{2x - (a + b)}{2\sqrt{(a - x)(x - b)}}$

Vedclass · Answer

(B) Let $u = \sqrt{a - x}$ and $v = \sqrt{x - b}$. Then $u^2 = a - x$ and $v^2 = x - b$.Note that $u^2 + v^2 = a - b$. The numerator is $u^3 - (b - x)v$. Since $b - x = -v^2$,the numerator is $u^3 + v^3$.Thus,$y = \frac{u^3 + v^3}{u + v} = u^2 - uv + v^2$.Substituting back,$y = (a - x) - \sqrt{(a - x)(x - b)} + (x - b) = a - b - \sqrt{-x^2 + (a + b)x - ab}$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2\sqrt{-x^2 + (a + b)x - ab}} \cdot (-2x + a + b)$.$\frac{dy}{dx} = \frac{2x - (a + b)}{2\sqrt{(a - x)(x - b)}}$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

If $y = \frac{(a - x)\sqrt{a - x} - (b - x)\sqrt{x - b}}{\sqrt{a - x} + \sqrt{x - b}}$,then $\frac{dy}{dx}$ wherever it is defined is equal to:

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