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Math Help - Derivative of a Parametric Function

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    Derivative of a Parametric Function

    Let x=t^2+t, and y=sin(t)

    Find d/dx(dy/dx)

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    \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

    \frac{d}{dx}\left[\frac{dy}{dx}\right] =  \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra  c{dx}{dt}}
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    I don't understand how that works out. I understand \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, but I don't understand this part:



    I don't have the intuition behind it.
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    Quote Originally Posted by nic42991 View Post
    I don't understand how that works out. I understand \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, but I don't understand this part:



    I don't have the intuition behind it.
    If you prefer, you can say that :

    \frac{df}{dx}=\frac{\frac{df}{dt}}{\frac{dx}{dt}} (1)

    Now, let f=\frac{dy}{dx} in (1), this gives :

    \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra  c{dx}{dt}}<br />

    And then transform \frac{dy}{dx}, by letting f=y in (1)
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  5. #5
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    let's look at it another way ...

    let \frac{dy}{dt} = y'(t) and \frac{dx}{dt} = x'(t)

    \frac{d}{dx}\left[\frac{y'(t)}{x'(t)}\right] =

    using the quotient and chain rule ...

    \frac{x'(t) \cdot y''(t) \cdot \frac{dt}{dx} - y'(t) \cdot x''(t) \cdot \frac{dt}{dx}}{[x'(t)]^2} =

    \frac{x'(t) \cdot y''(t) - y'(t) \cdot x''(t)}{[x'(t)]^2 \cdot \frac{dx}{dt}} =

    \frac{x'(t) \cdot y''(t) - y'(t) \cdot x''(t)}{[x'(t)]^3} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra  c{dx}{dt}}
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