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  1. #1
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    expression manipulating

    I have done the first part and found the new limits but I can't do the last part.
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  2. #2
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    Re: expression manipulating

    Quote Originally Posted by Stuck Man View Post
    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*}
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  3. #3
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    Re: expression manipulating

    I've done that part.
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    Re: expression manipulating

    Quote Originally Posted by Stuck Man View Post
    I've done that part.
    You're welcome ><
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  5. #5
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    Re: expression manipulating

    \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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