I'm trying to get back into maths after a couple of years out.
Looking at the opening chapter of a text book on Nonlinear Ordinary Differential Equations (the context is somewhat irrelevant) I've found an equation that I cannot make sense of:
$\ddot{x} = \frac{\mathrm d \dot{x}}{\mathrm d t} = \frac{\mathrm d \dot{{x}}}{\mathrm d x} \frac{\mathrm d x}{\mathrm d t} = \frac{\mathrm d}{\mathrm d x} (\frac{1}{2} \dot{x}^2)$
I undertand it using the chain rule, and it's clear that one of the $\dot{x}$'s comes from the $\frac{\mathrm d x}{\mathrm d t}$. However what I can't understand is how $\frac{\mathrm d \dot{x}}{\mathrm d x}$ becomes $\frac{1}{2} \dot{x}$.
I realise this is probably something obvious, but like I say, it's been a while.
You forgot to take into account the $\frac{\mathrm d}{\mathrm d x}$ in front of the bracket. Using the product rule on the last expression, we get $$ \frac{\mathrm d}{\mathrm d x} \left(\frac{1}{2} \dot{x}^2\right) = \frac{1}{2}\frac{\mathrm d}{\mathrm d x} \left( \dot{x}^2\right) \\ = \frac{1}{2}\left( \frac{\mathrm d\dot x}{\mathrm d x} \dot{x} + \dot x\frac{\mathrm d\dot x}{\mathrm d x}\right)\\ = \frac{\mathrm d\dot x}{\mathrm d x} \dot{x} = \frac{\mathrm d\dot x}{\mathrm d x} \frac{\mathrm dx}{\mathrm d t} $$ Your textbook just did this in reverse.