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How do I differentiate these parametric functions? FP1 watch

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    y = 2a^{1/2}x^{1/2}

    y = \frac{c^2}{x}
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    For the first multiply by power of x then take 1 away from power

    For second rewrite as this

    y=c^2x^{-1}

    Then try differentiating.
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    (Original post by raiden95)
    For the first multiply by power of x then take 1 away from power

    For second rewrite as this

    y=c^2x^{-1}
    y = -c^2x^{-2}

    So this? y = \frac{-c^2}{x^2}
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    Differentiate normally leaving the constants a and c unchanged.
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    (Original post by lizz-ie)
    y = -c^2x^{-2}

    So this? y = \frac{-c^2}{x^2}
    Yep
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    (Original post by lizz-ie)
    y = -c^2x^{-2}

    So this? y = \frac{-c^2}{x^2}
    \frac{dy}{dx} = \frac{-c^2}{x^2}

    Yes, but don't forget the dy/dx
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    (Original post by Lunch_Box)
    Differentiate normally leaving the constants a and c unchanged.
    So y = a^{1/2}x^{-1/2} ?
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    (Original post by SheldonWannabe)
    If you multiplied the whole thing by x and took away the power, you would be left with the same thing, just with yx on the left hand side. So that is wrong...
    you simply move the power of x to the coefficient and minus one from the power of x, just differentiate like you would with any other equation.
    That is exactly what I mean ? The OP clearly understood.

    I didn't say multiply whole thing by x, i said multiply by power of x
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    (Original post by raiden95)
    \frac{dy}{dx} = \frac{-c^2}{x^2}

    Yes, but don't forget the dy/dx
    Thank you so much!
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    (Original post by lizz-ie)
    So y = a^{1/2}x^{-1/2} ?
    Fast learner
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    (Original post by lizz-ie)
    So y = a^{1/2}x^{-1/2} ?
    Yes thats correct but with dy/dx
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    (Original post by raiden95)
    That is exactly what I mean ? The OP clearly understood.

    I didn't say multiply whole thing by x, i said multiply by power of x
    Haha oh damn yeah, I read that completely wrong... Now I feel stupid
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    Just a little note, these are not parametric equations. They are Cartesian equations of the parabola and the rectangular hyperbola respectively.
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    (Original post by Ateo)
    Just a little note, these are not parametric equations. They are Cartesian equations of the parabola and the rectangular hyperbola respectively.
    I was also confused about that.
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    (Original post by Ateo)
    Just a little note, these are not parametric equations. They are Cartesian equations of the parabola and the rectangular hyperbola respectively.
    If we're being really pedantic, the first equation gives the upper branch of the parabola because of the convention that we take positive square roots to define a function.

    The alternative form y^2 = 4ax gives both branches of the parabola.
 
 
 
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Updated: April 6, 2013
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