# How to evaluate $\frac{\partial}{\partial\theta}(\frac{d\theta}{dt})$ and $\frac{d}{d\theta}(\frac{d\theta}{dt})$

This question actually occurred in a Physics problem, however I thought it would be more appropriate to ask it here as it's only about the mathematical part.

The question was about a pendulum with an accelerated support point using Lagrangian mechanics. The Lagrangian involved the following term:

$$L=\dots+mlsin(\theta)\dot\theta\dot{y}$$

where $\dot{y}=\frac{1}{2}at^2$. When evaluating the RHS of the Euler-Lagrange equation $\frac{d}{dt}(\frac{\partial{L}}{\partial\dot\theta})=\frac{\partial{L}}{\partial\theta}$ I had the following problem:

How do I evaluate $\frac{\partial}{\partial\theta}(\frac{d\theta}{dt})$?

My instinct would tell me as $\theta$ is a function of t only that this would be equivalent to evaluating $\frac{d}{d\theta}(\frac{d\theta}{dt})$. Using the chani rule:

$\frac{d}{d\theta}(\frac{d\theta}{dt})$=$\frac{d}{dt}(\frac{d\theta}{dt})\frac{dt}{d\theta}$=$\frac{\ddot\theta}{\dot\theta}$

Therefore given the whole differentiated term using the product rule as:

$mlcos(\theta)\dot\theta\dot{y}+mlsin(\theta)\frac{\ddot\theta}{\dot\theta}\dot{y}$

However my textbook leaves out the second term and only takes the first term, i.e. it takes

$\frac{\partial}{\partial\theta}(\frac{d\theta}{dt})=0$

Summarised my question is: How do I evaluate these two expressions:

$\frac{\partial}{\partial\theta}(\frac{d\theta}{dt})$ and $\frac{d}{d\theta}(\frac{d\theta}{dt})$

when $\theta=\theta(t)$

In the Euler–Lagrange equations, when computing $\partial L/\partial \theta$ and $\partial L/\partial \dot\theta$, the quantities $\theta$ and $\dot \theta$ are treated as independent variables, so $\frac{\partial}{\partial \theta} \dot \theta = 0$. At this stage, you shouldn't think of $\theta$ as depending on $t$. (That doesn't enter until you compute $\frac{d}{dt} (\partial L/\partial \dot\theta)$.)