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# If , y=sqrt(x/a)-sqrt(a/x) So, Prove that ?? 2xy(dy/dx)= x/a-a/x

## If , $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$ So, Prove that ? $2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{a} - \frac{a}{x}$

Jun 17, 2018

Given: $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$

Differentiating:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{\frac{a}{x}} + \sqrt{\frac{x}{a}}}{2 x}$

Multiply both sides by2xy:

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x y \frac{\sqrt{\frac{a}{x}} + \sqrt{\frac{x}{a}}}{2 x}$

$\frac{2 x}{2 x}$ becomes 1:

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\sqrt{\frac{a}{x}} + \sqrt{\frac{x}{a}}\right)$

Substitute $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$ on the right side only:

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}\right) \left(\sqrt{\frac{a}{x}} + \sqrt{\frac{x}{a}}\right)$

Please observe that the right side is the pattern $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$ where $a = \sqrt{\frac{x}{a}}$ and $b = \sqrt{\frac{a}{x}}$

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{a} - \frac{a}{x} \text{ Q.E.D.}$

Jun 17, 2018

Kindly see a Proof in Explanation.

#### Explanation:

We have, $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$.

Diff.ing w.r.t. $x , \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{a}} \cdot \left(\frac{1}{2} {x}^{- \frac{1}{2}}\right) - \sqrt{a} \cdot \left(- \frac{1}{2} {x}^{- \frac{3}{2}}\right)$.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{a x}} + \frac{\sqrt{a}}{2 x \sqrt{x}}$.

Multiplying by $2 x , 2 x \frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}$.

Again multiplying by $y , 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\right)$,

$i . e . , 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}\right) \left(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\right)$.

$\Rightarrow 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{a} - \frac{a}{x}$, as desired!

Jun 17, 2018

We seek to show that:

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{a} - \frac{a}{x}$ where $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$

Squaring the expression, and then expanding, we get:

${y}^{2} = {\left(\sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}\right)}^{2}$
$\setminus \setminus \setminus \setminus = {\left(\sqrt{\frac{x}{a}}\right)}^{2} - 2 \left(\sqrt{\frac{x}{a}}\right) \left(\sqrt{\frac{a}{x}}\right) + {\left(\sqrt{\frac{a}{x}}\right)}^{2}$
$\setminus \setminus \setminus \setminus = \frac{x}{a} - 2 + \frac{a}{x}$

Differentiating Implicitly, we get

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{a} - \frac{a}{x} ^ 2$

Finally, multiplying by $x$ we get the desired result:

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{a} - \frac{a}{x} \setminus \setminus \setminus$ QED