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I have a function $y(x)$ defined via a parameter as $y(t)$ and x(t). Using a common convention of the dot for $\dfrac{d}{dt}$ and prime for $\dfrac{d}{dx}$, I understand that:

$$\dot{y}=y'\dot{x}$$

I need a similar formula for the second derivative, but I am having a hard time finding it. My own math is a bit rusty, so this is what I've been able to figure:

$$\ddot{y}=y''\dot{x}^2+y'\ddot{x}$$

If this is off, what is the correct formula? Sorry for a trivial question and thanks for your help!

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We have $y=y(x(t))$ \begin{eqnarray*} \frac{dy}{dt} =\frac{dy}{dx} \frac{dx}{dt} \end{eqnarray*} and we can write this as $\dot{y}=y'\dot{x}$. Now differnatiate this \begin{eqnarray*} \frac{d^2y}{dt^2} = \frac{d}{dt} \left( \frac{dy}{dx} \right) \frac{dx}{dt}+ \frac{dy}{dx} \frac{d^2x}{dt^2} \end{eqnarray*} So we need \begin{eqnarray*} \frac{d}{dt} \left( \frac{dy}{dx} \right) =\frac{dx}{dt} \frac{d}{dx} \left( \frac{dy}{dx} \right). \end{eqnarray*} So \begin{eqnarray*} \frac{d^2y}{dt^2} = \frac{d^2y}{dx^2} \left( \frac{dx}{dt}\right)^2+ \frac{dy}{dx} \frac{d^2x}{dt^2} \end{eqnarray*} and we have $\ddot{y}=y''(\dot{x})^2+y'\ddot{x} $.

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    $\begingroup$ Thanks so much! It is indeed the part in the middle that I wanted to confirm: $\dfrac{d}{dt}=\dfrac{dx}{dt}\dfrac{d}{dx}$ $\endgroup$ – safesphere Nov 22 '17 at 0:45

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