Question #60b11

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Question #60b11

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Answer 1 · 2017-11-16T00:08:17

When $y = \sqrt{\frac{x}{m}} + \sqrt{\frac{m}{x}}$ $y=sqrt(x/m)+sqrt(m/x)$ , we observe that $2 x y \frac{d y}{d x} - \frac{x}{m} + \frac{m}{x} = 0$ $2xydy/dx-x/m+m/x=0$ .

Explanation:

$y = \sqrt{\frac{x}{m}} + \sqrt{\frac{m}{x}}$ $y=sqrt(x/m)+sqrt(m/x)$

Then I assume you're asking, what's the expression $2 x y \frac{d y}{d x} - \frac{x}{m} + \frac{m}{x}$ $2xydy/dx-x/m+m/x$ equivalent to?

Well, we need to find $\frac{d y}{d x}$ $dy/dx$ . First let's rewrite $y$ $y$ .

$y = m^{- \frac{1}{2}} x^{\frac{1}{2}} + m^{\frac{1}{2}} x^{- \frac{1}{2}}$ $y=m^(-1/2)x^(1/2)+m^(1/2)x^(-1/2)$

So now we can differentiate with the power rule:

$\frac{d y}{d x} = \frac{1}{2} m^{- \frac{1}{2}} x^{- \frac{1}{2}} - \frac{1}{2} m^{\frac{1}{2}} x^{- \frac{3}{2}} = \frac{1}{2 \sqrt{m x}} - \frac{\sqrt{m}}{2 x \sqrt{x}}$ $dy/dx=1/2m^(-1/2)x^(-1/2)-1/2m^(1/2)x^(-3/2)=1/(2sqrt(mx))-sqrtm/(2xsqrtx)$

So then we want to know about:

$2 x y \frac{d y}{d x} - \frac{x}{m} + \frac{m}{x}$ $2xydy/dx-x/m+m/x$

$= 2 x (\sqrt{\frac{x}{m}} + \sqrt{\frac{m}{x}}) (\frac{1}{2 \sqrt{m x}} - \frac{\sqrt{m}}{2 x \sqrt{x}}) - \frac{x}{m} + \frac{m}{x}$ $=2x(sqrt(x/m)+sqrt(m/x))(1/(2sqrt(mx))-sqrtm/(2xsqrtx))-x/m+m/x$

FOIL the two binomials:

$= 2 x (\frac{\sqrt{x}}{\sqrt{m}} \frac{1}{2 \sqrt{m} \sqrt{x}} - \frac{\sqrt{x}}{\sqrt{m}} \frac{\sqrt{m}}{2 x \sqrt{x}} + \frac{\sqrt{m}}{\sqrt{x}} \frac{1}{2 \sqrt{m} \sqrt{x}} - \frac{\sqrt{m}}{\sqrt{x}} \frac{\sqrt{m}}{2 x \sqrt{x}}) - \frac{x}{m} + \frac{m}{x}$ $=2x(sqrtx/sqrtm1/(2sqrtmsqrtx)-sqrtx/sqrtmsqrtm/(2xsqrtx)+sqrtm/sqrtx1/(2sqrtmsqrtx)-sqrtm/sqrtxsqrtm/(2xsqrtx))-x/m+m/x$

And simplify:

$= 2 x (\frac{1}{2 m} - \frac{1}{2 x} + \frac{1}{2 x} - \frac{m}{2 x^{2}}) - \frac{x}{m} + \frac{m}{x}$ $=2x(1/(2m)-1/(2x)+1/(2x)-m/(2x^2))-x/m+m/x$

Distribute:

$= (\frac{x}{m} - \frac{m}{x}) - \frac{x}{m} + \frac{m}{x}$ $=(x/m-m/x)-x/m+m/x$

$= 0$ $=0$

Cool!

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Question #60b11

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