Simple derivate question

Question

In a paper I am reading about dynamics systems, they set the following variables:

$a(\theta) = \ddot{\theta}$, $b(\theta) = \dot{\theta}^2$

Where $\dot{\theta}$ and $\ddot{\theta}$ are the first and second derivate respecting time.

Then, they claim that $a(\theta)$ and $b(\theta)$ are related as follows:

$\dot{b}(\theta) = b'(\theta)\dot{\theta} = \frac{d(\dot{\theta}^2)}{dt} = 2\ddot{\theta}\dot{\theta} = 2a(\theta)\dot{\theta}$

Where $b'(\theta)$ is the derivative of $b(\theta)$ with respect to $\theta$.

In the above expression, I understand that $\dot{b}(\theta) = b'(\theta)\dot{\theta}$ because of the chain rule. But I cannot see why $b'(\theta)\dot{\theta} = \frac{d(\dot{\theta}^2)}{dt}$.

Any help is appreciated. Thank you!

Answer 1

$\frac{d(\dot\theta^2)}{dt}$ is not obtained from $b'(\theta)\dot\theta$, but from $b(\theta)=\dot\theta^2$, by derivation.

Answer 2

You have $b(\theta) = \dot{\theta}^2$ .......(1)

Where $\dot{\theta}$ and $\ddot{\theta}$ are the first and second derivate respecting time.

In the above expression, you understand that $\dot{b}(\theta) = b'(\theta)\dot{\theta}$ because of the chain rule. Then $b'(\theta)\dot{\theta} = \dot{b}(\theta)= frac {d(b \theta}{dt}= \frac{d(\dot{\theta}^2)}{dt}$.[by 1]

Answer 3

The chain rule is your friend $$ \dfrac{d}{dt}b(\theta) = \dfrac{db}{d\theta}\cdot \dfrac{d\theta}{dt} $$ but you also have $b=\dot{\theta}^2$