Chain rule for partial derivatives.

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Is it true that:

$$\frac{1}{2}\frac{d\mu}{dq}\cdot\frac{p^2}{\mu^2}=\frac{1}{2}\frac{d\mu}{dq}\cdot\frac{\partial}{\partial \mu}\left(\frac{-p^2}{2\mu}\right)=-\frac{\partial}{\partial q}\left(\frac{p^2}{2\mu}\right),\;\;\; \mu=\mu(q)$$

And if not why? I've seen the first equation being equal to the last part, but the middle step is my interpretation of how they are equal. My reasoning is that $\dfrac{d\mu}{dq}=\dfrac{\partial \mu}{\partial q}$ as $\mu$ is just a function of $q$. And so I can use chain rule for partials. Is this correct? (By the way it's for Hamiltonian equations).

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1st and 3rd => you know

1st and 2nd => yeah just apply the second operator

2nd and 3rd => yeah just cancel the $d \mu$ with $\partial \mu$

All the above assumes the middle term has an extra 2.