If , $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$ $y=sqrt(x/a)-sqrt(a/x)$ So, Prove that ?? $2 x y (\frac{d y}{d x}) = \frac{x}{a} - \frac{a}{x}$ $2xy(dy/dx)= x/a-a/x$

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If , $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$ $y=sqrt(x/a)-sqrt(a/x)$ So, Prove that ?? $2 x y (\frac{d y}{d x}) = \frac{x}{a} - \frac{a}{x}$ $2xy(dy/dx)= x/a-a/x$

If , $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$ $y=sqrt(x/a)-sqrt(a/x)$ So, Prove that $?$ $?$
$2 x y \frac{d y}{d x} = \frac{x}{a} - \frac{a}{x}$ $2xydy/dx= x/a-a/x$

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Answer 1 · 2018-06-17T17:47:17

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

Differentiating:

$\frac{d y}{d x} = \frac{\sqrt{\frac{a}{x}} + \sqrt{\frac{x}{a}}}{2 x}$ $dy/dx = (sqrt(a/x) + sqrt(x/a))/(2 x)$

Multiply both sides by2xy:

$2 x y \frac{d y}{d x} = 2 x y \frac{\sqrt{\frac{a}{x}} + \sqrt{\frac{x}{a}}}{2 x}$ $2xydy/dx = 2xy(sqrt(a/x) + sqrt(x/a))/(2 x)$

$\frac{2 x}{2 x}$ $(2x)/(2x)$ becomes 1:

$2 x y \frac{d y}{d x} = y (\sqrt{\frac{a}{x}} + \sqrt{\frac{x}{a}})$ $2xydy/dx = y(sqrt(a/x) + sqrt(x/a))$

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

$2 x y \frac{d y}{d x} = (\sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}) (\sqrt{\frac{a}{x}} + \sqrt{\frac{x}{a}})$ $2xydy/dx = (sqrt(x/a)-sqrt(a/x))(sqrt(a/x) + sqrt(x/a))$

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

$2 x y \frac{d y}{d x} = \frac{x}{a} - \frac{a}{x} Q.E.D.$ $2xydy/dx = x/a-a/x" Q.E.D."$

Answer 2 · 2018-06-17T18:19:21

Kindly see a Proof in Explanation.

Explanation:

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

Diff.ing w.r.t. $x, \frac{d y}{d x} = \frac{1}{\sqrt{a}} \cdot (\frac{1}{2} x^{- \frac{1}{2}}) - \sqrt{a} \cdot (- \frac{1}{2} x^{- \frac{3}{2}})$ $x, dy/dx=1/sqrta*(1/2x^(-1/2))-sqrta*(-1/2x^(-3/2))$ .

$:. dy/dx=1/(2sqrt(ax))+sqrta/(2xsqrtx)$ .

Multiplying by $2x, 2xdy/dx=sqrt(x/a)+sqrt(a/x)$ .

Again multiplying by $y, 2xydy/dx=y(sqrt(x/a)+sqrt(a/x))$ ,

$i.e., 2xydy/dx=(sqrt(x/a)-sqrt(a/x))(sqrt(x/a)+sqrt(a/x))$ .

$rArr 2xydy/dx=x/a-a/x$ , as desired!

Answer 3 · 2018-06-17T19:59:19

We seek to show that:

$2xydy/dx= x/a-a/x$ where $y=sqrt(x/a)-sqrt(a/x)$

Squaring the expression, and then expanding, we get:

$y^2 = (sqrt(x/a)-sqrt(a/x))^2$
$\ \ \ \= (sqrt(x/a))^2 - 2(sqrt(x/a))(sqrt(a/x)) + (sqrt(a/x))^2$
$\ \ \ \= x/a - 2 + a/x$

Differentiating Implicitly, we get

$2y dy/dx = 1/a -a/x^2$

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

$2xy dy/dx = x/a -a/x \ \ \$ QED

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If , $y = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}$ $y=sqrt(x/a)-sqrt(a/x)$ So, Prove that ?? $2 x y (\frac{d y}{d x}) = \frac{x}{a} - \frac{a}{x}$ $2xy(dy/dx)= x/a-a/x$