I am reading this paper where it is given
where $x = g(s,\alpha)$ and $u$ is a function of $x$.
The above expression is for a given $\alpha$.
The first equation is straight forward from chain rule. However, I am not able to obtain the second equation.
When I apply chain rule, this is what I get
$$\frac{d^2u}{ds^2} = \frac{d}{ds}\Big(\frac{du}{ds}\Big) = \frac{d}{dx}\Big(\frac{du}{dx}\ g'\Big)\frac{dx}{ds}$$ $$\frac{d^2u}{ds^2} = \bigg[g' \frac{d}{dx}\Big(\frac{du}{dx}\Big) + \frac{d}{dx}\Big(\frac{dx}{ds}\Big) \frac{du}{dx}\bigg]g'$$
where $g' = \frac{dx}{ds}$
How do I go further to arrive at the above expression?
First notice that by chain rule we have $$\dfrac d{dx} = \frac 1{g'(s,\alpha)}\dfrac d{ds}$$ and thus we have
\begin{align}\dfrac{d^2u}{dx^2} &= \dfrac d{dx}\left(\dfrac{du}{dx}\right)= \frac 1{g'(s,\alpha)}\dfrac d{ds}\left(\frac 1{g'(s,\alpha)}\dfrac {du}{ds}\right)\\ &= \frac 1{g'(s,\alpha)}\left(\dfrac d{ds}\left(\frac 1{g'(s,\alpha)}\right)\dfrac {du}{ds}+\frac 1{g'(s,\alpha)}\dfrac{d^2u}{ds^2}\right)\\ &= \frac 1{g'(s,\alpha)}\left(\dfrac {-g''(s,\alpha)}{g'(s,\alpha)^2}\dfrac {du}{ds}+\frac 1{g'(s,\alpha)}\dfrac{d^2u}{ds^2}\right).\end{align}