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If, y = (√x)+(1/√x) so prove that ? 2x(dy/dx)+y = 2√x

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Explanation

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Explanation:

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2
Jun 17, 2018

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

Explanation:

$y = \sqrt{x} + \frac{1}{\sqrt{x}}$

$\sqrt{x} . y = x + 1$

$\left(\frac{1}{2 \cdot \sqrt{x}}\right) . y + \sqrt{x} . \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$y = 2. \sqrt{x} \left(1 - \sqrt{x} . \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

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

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

Then teach the underlying concepts
Don't copy without citing sources
preview
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Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

1
Jun 19, 2018

Please see a Proof in the Explanation.

Explanation:

Given that, $y = \sqrt{x} + \frac{1}{\sqrt{x}} = {x}^{\frac{1}{2}} + {x}^{- \frac{1}{2}}$.

Recall that, $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n \cdot {x}^{n - 1}$.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \cdot {x}^{\frac{1}{2} - 1} + \left(- \frac{1}{2}\right) \cdot {x}^{- \frac{1}{2} - 1}$.

$= \frac{1}{2} \cdot {x}^{- \frac{1}{2}} - \frac{1}{2} \cdot {x}^{- \frac{3}{2}}$,

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left\{{x}^{- \frac{1}{2}} - {x}^{- \frac{3}{2}}\right\}$.

Multiplying this eqn. by $2 x$, we get,

$2 x \frac{\mathrm{dy}}{\mathrm{dx}} = \cancel{2} x \cdot \frac{1}{\cancel{2}} \left\{{x}^{- \frac{1}{2}} - {x}^{- \frac{3}{2}}\right\}$,

$= x \cdot {x}^{- \frac{1}{2}} - x \cdot {x}^{- \frac{3}{2}}$,

$i . e . , 2 x \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\frac{1}{2}} - {x}^{- \frac{1}{2}} = \sqrt{x} - \frac{1}{\sqrt{x}}$.

Finally, adding $y = \sqrt{x} + \frac{1}{\sqrt{x}}$, we have,

$2 x \frac{\mathrm{dy}}{\mathrm{dx}} + y = \left(\sqrt{x} \cancel{- \frac{1}{\sqrt{x}}}\right) + \left(\sqrt{x} \cancel{+ \frac{1}{\sqrt{x}}}\right)$,

$\mathmr{and} , 2 x \frac{\mathrm{dy}}{\mathrm{dx}} + y = 2 \sqrt{x}$,

as Respected Abhishek Malviya has readily derived!

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