expression manipulating

**Stuck Man** · November 17th 2011, 05:43 AM

I have done the first part and found the new limits but I can't do the last part.

**Prove It** · November 17th 2011, 05:49 AM

Originally Posted by Stuck Man

I have done the first part and found the new limits but I can't do the last part.

$\displaystyle \begin{align*} y &= \frac{x}{1 + x^2} \\ \frac{y}{x} &= \frac{1}{1 + x^2} \\ \frac{x}{y} &= 1 + x^2 \\ \left(\frac{x}{y}\right)^2 &= (1 + x^2)^2 \\ \left(\frac{x}{y}\right)^2 &= 1 + 2x^2 + x^4 \\ \left(\frac{x}{y}\right)^2 - 2x^2 &= 1 + x^4 \end{align*}$

**Stuck Man** · November 17th 2011, 06:23 AM

I've done that part.

**Prove It** · November 17th 2011, 06:40 PM

Originally Posted by Stuck Man

I've done that part.

You're welcome ><

**tom@ballooncalculus** · November 18th 2011, 02:48 AM

$\displaystyle{y = \frac{x}{1 + x^2}}$

And,

$\displaystyle{\frac{dy}{dx} = \frac{1 - x^2}{(1 + x^2)^2}}$

Therefore,

$\displaystyle{\frac{1 - x^2}{(1 + x^2) \sqrt{1 + x^4}}}$

$=\displaystyle{\frac{1 - x^2}{(1 + x^2) \sqrt{(\frac{x}{y})^2 - 2x^2}}}$

$=\displaystyle{\frac{1 - x^2}{(1 + x^2) \sqrt{(\frac{x}{y})^2 (1 - 2y^2)}}}}$

$=\displaystyle{\frac{1 - x^2}{(1 + x^2)\ \frac{x}{y} \sqrt{1 - 2y^2}}}$

$=\displaystyle{\frac{1 - x^2}{x(1 + x^2)}\ \frac{y}{\sqrt{1 - 2y^2}}}$

$=\displaystyle{\frac{1 + x^2}{x}\ \frac{y}{\sqrt{1 - 2y^2}}\ \frac{1 - x^2}{(1 + x^2)^2}}}$

$=\displaystyle{\frac{1}{y}\ \frac{y}{\sqrt{1 - 2y^2}}\ \frac{dy}{dx}}}$

$=\displaystyle{\frac{1}{\sqrt{1 - 2y^2}}\ \frac{dy}{dx}}}$

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