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Thread: Derivative of a Parametric Function

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    Derivative of a Parametric Function

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

    Find $\displaystyle d/dx(dy/dx)$

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

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



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

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

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

    $\displaystyle \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra c{dx}{dt}}
    $

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

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

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

    using the quotient and chain rule ...

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

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

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