Confusion in chain rule for $y^{''}$

Question

Currently outside so there's no way to type latex out. Hence I'll attach the picture here. What I don't get is how they use chain rule to get to this answer. For chain rule wise, I know that if $y=f(u) ,u=g(x)$ then we have $dy/dx = dy/du \cdot du/dx$. Anyone can enlighten me on how the answer works in details?

Answer 1

Since the $y'$ may be distracting you, let us define $\def\d{\mathrm d}z=\dfrac{\d y}{\d x}$

$\begin{align} \dfrac{\d ^2 y}{\d x~^2} ~&= \dfrac{\d~}{\d x}\dfrac{\d y}{\d x} \\ &= \dfrac{\d z}{\d x} & & \text{substituting} \\ &= \dfrac{\d y}{\d x}\dfrac{\d z}{\d y} &&\text{chain rule} \\ &= z\dfrac{\d z}{\d y} && \text{substituting}\\ &= \dfrac{\d(\tfrac 12 z^2)}{\d y\hspace{5ex}} && \text{since } \dfrac{\d z^2}{\d y}= 2 z\dfrac{\d z}{\d y} \\ &=\dfrac 12\dfrac{\d (y')^2}{\d y\hspace{4ex}} && \text{because }z=y' \end{align}$

Answer 2

Note that its $\frac {d}{dy} $ in Rhs . So multiply and divide by $dx $ to get $\frac {d}{dx}(\frac {y'^2}{2})\frac {dx}{dy}=\frac {y'y"}{y'} =y"$ (assuming $y'\neq 0$)

Answer 3

It is quite nonrigorous to proceed like that!

Better, and in fact simpler, to note that $y''y'=F(y)y'$ and so, integrating with respecto to $t$, $$ \frac12(y')^2=G(y), $$ where $G$ is a primitive of $F$.