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Math Help - Differentiating Implicitly!

  1. #1
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    Differentiating Implicitly!

    Find dy/dx. Assume b, p, and m are constants.

    bx^2 -py^2 = m^2

    i need some help with this, i know you cant differentiate constants, so how would i do this one...

    mathlete
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  2. #2
    Super Member wingless's Avatar
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    For example, try to differentiate 3y^2

    \frac{d}{dx}3y^2

    \frac{d}{dx}\cdot\frac{dy}{dy}3y^2

    \frac{dy}{dx}6y

    6yy'

    We can generalize this to,
    \frac{d}{dx}f(y) = \frac{d}{dx}\frac{dy}{dy}f(y) = y'\cdot f'(x)

    So,

    bx^2 - p y^2 = m^2

    2b x - 2p y y' = 0

    Take y' from here..
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  3. #3
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    Quote Originally Posted by mathlete View Post
    Find dy/dx. Assume b, p, and m are constants.

    bx^2 -py^2 = m^2

    i need some help with this, i know you cant differentiate constants, so how would i do this one...

    mathlete
    You can differentiate constants - they are just zero!

    bx^2 -py^2 = m^2

    Derive y with respect to x using implicit differentiation. ..

    bx^2 -py^2 = m^2 \Rightarrow 2bx - 2py \frac{dy}{dx} = 0
    py\frac{dy}{dx} = bx
    \frac{dy}{dx} = \frac{bx}{py}

    Substitute y back into the equation and you are golden!

    bx^2 -py^2 = m^2
    py^2 = bx^2 - m^2
    y^2 = \frac{bx^2 - m^2}{p}
    y = \pm \sqrt{\frac{bx^2 - m^2}{p}}

    \frac{dy}{dx} = \frac{bx}{\pm p\sqrt{\frac{bx^2 - m^2}{p}}}

    Let's check out work another way...

    bx^2 -py^2 = m^2

    Solve for y...

    y = \pm \left(\frac{bx^2 - m^2}{p}\right)^{0.5}

    Take the derivative with respect to x

    y' = \pm \frac{2bx}{2p}\left(\frac{bx^2 - m^2}{p}\right)^{-0.5} \Rightarrow \pm \frac{bx}{p \sqrt{\frac{bx^2 - m^2}{p}}}
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