If $\dfrac{d\theta}{dt} = \omega$ then show that $\dfrac{d^2 \theta}{dt^2} = \omega \dfrac{d\omega}{d\theta}$.

Question

Let $$ \frac{d\theta}{dt} = \omega$$ show that $\frac{d^2\theta}{dt^2}$ can be expressed as $$\omega\frac{d\omega}{d\theta}$$

My approach is as follows: $$\frac{d^2\theta}{dt^2} = \frac{d}{dt}\biggl(\frac{d\theta}{dt}\biggl)$$ if we substitute $dt=\frac{d\theta}{\omega}$ then $$\frac{d^2\theta}{dt^2} = \frac{d}{\frac{d\theta}{\omega}}\biggl(\frac{d\theta}{\frac{d\theta}{\omega}}\biggl) = \frac{d(\omega)}{d\theta}\biggl({\frac{d\theta}{d\theta}\omega}\biggl) = \omega\frac{d\omega}{d\theta}$$

However, I have a feeling that this approach is mathematically incorrect. Hence I was wondering

what other approaches are there, with mathematical rigor.

Moreover, when changing variables of derivatives like this, be it first or second derivatives, are

there any general rules that apply?

Answer 1

You can express $\omega$ as a composite function $\omega=f(\theta(t))$

than use the chain rule

$$\frac{d \omega}{d t}=\frac{d \omega}{d \theta}\frac{d \theta}{d t}=\omega\frac{d \omega}{d \theta}$$

Answer 2

$$\begin{array}{rcl} \dfrac{\mathrm d^2\theta}{\mathrm dt^2} &=& \dfrac{\mathrm d}{\mathrm dt} \dfrac{\mathrm d\theta}{\mathrm dt} \\ &=& \dfrac{\mathrm d\omega}{\mathrm dt} \\ &=& \dfrac{\mathrm d\omega}{\mathrm d\theta} \dfrac{\mathrm d\theta}{\mathrm dt} \\ &=& \dfrac{\mathrm d\theta}{\mathrm dt} \dfrac{\mathrm d\omega}{\mathrm d\theta} \\ &=& \omega \dfrac{\mathrm d\omega}{\mathrm d\theta} \\ \end{array}$$