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Math Help - Calculus, dervatives

  1. #1
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    Calculus, dervatives

    Hi, I am finding it difficult to undertake this question, can anyone please assist me?

    Find dy/dx in terms of x and y for:

    x^2 = arctan(y) / 1+y^2



    Any help much appreciated,

    Dranalion.
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  2. #2
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    Quote Originally Posted by Dranalion View Post
    Hi, I am finding it difficult to undertake this question, can anyone please assist me?

    Find dy/dx in terms of x and y for:

    x^2 = arctan(y) / 1+y^2



    Any help much appreciated,

    Dranalion.
    The derviative of x^2 is, of course, 2x. By the quotient rule, the derivative of f(y)/g(y) is (f'g- fg')/g. Further, the derivative of arctan(y) is \frac{1}{1+ y^2} so the derivative of \frac{arctan(y)}{1+ y^2}, with respect to y, is \frac{\frac{1}{1+ y^2}(1+y^2)- arctan(y)(2y)}{(1+ y^2)^2}= \frac{1- y arctan(y)}{(1+ y^2)^2}.
    The derivative with respect to x, rather than y, is that times \frac{dy}{dx} so we have 2x= \frac{1- y arctan(y)}{(1+ y^2)^2}\frac{dy}{dx}.
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  3. #3
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    I got this however, what have I done?

    (1+y^2)*x^2 = arctan(y)
    2x(1+y^2) + x^2(1+2yy') = y'/(1+y^2)
    2x+2xy^2+x^2+2yy'x^2 = y'/(1+y^2)
    2x+2xy^2+x^2 = y'/(1+y^2) - 2yy'x^2 = y'(1/(1+y^2) - 2yx^2)
    y' = (2x+2xy^2+x^2) / (1/(1+y^2) - 2yx^2)
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  4. #4
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    Hello, Dranalion!

    Find \frac{dy}{dx} for: . x^2 \:=\: \frac{\arctan(y)}{1+y^2}
    First, I'd get rid of the quotient . . . x^2(1+y^2) \:=\:\arctan(y)<br />

    And might as well get rid of the product . . . x^2 + x^2y^2 \:=\:\arctan(y)


    Differentiate: . 2x + x^2\!\cdot\!2y\!\cdot\!y' + 2xy^2 \:=\:\frac{y'}{1+y^2} \quad\Rightarrow\quad 2x^2y\!\cdot\!y' + 2x(1 + y^2) \:=\:\frac{y'}{1+y^2}

    . . 2x^2y\!\cdot\!y'(1+y^2) + 2x(1+y^2)^2 \:=\:y' \quad\Rightarrow\quad y' - 2x^2y(1+y^2)\!\cdot\!y' \:=\:2x(1+y^2)^2


    Factor: . \left[1 - 2x^2y(1+y^2)\right]\!\cdot\!y' \:=\:2x(1+y^2)^2 \quad\Rightarrow\quad y' \;=\;\frac{2x(1+y^2)^2}{1-2x^2y(1+y^2)}

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  5. #5
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    I got \frac{-(1+y^2)}{2x}
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