Let $f$ be defined on $[a,b]$ and g a continuous function defined on $[\alpha , \beta ]$ that is differentiable at $\gamma \in (\alpha, \beta)$ with $g(\gamma)=c\in(a,b)$. Show that if $g'(\gamma)>0$ then $D^+(f\circ g)(\gamma)=D^+f(c)\cdot g'(\gamma)$ where $D^+=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f(x+t)-f(x)}{t}\right]$
Heres some work I did, I was wondering if I am working in the right direction.
$$ \begin{align} D^+(f\circ g)(\gamma)&=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f\circ g(\gamma+t)-f\circ g(\gamma)}{t}\right] \\ &=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f( g(\gamma+t))-f( g(\gamma))}{t}\right] \\ &= \lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f( g(\gamma+t))-f( g(\gamma))}{g(\gamma+t)-g(\gamma)}\cdot \dfrac{g(\gamma+t)-g(\gamma)}{t}\right] \\ &(?)= \lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\left( \dfrac{f( g(\gamma+t))-f( g(\gamma))}{g(\gamma+t)-g(\gamma)} \right) \cdot \sup_{0<|t|\le h} \left( \dfrac{g(\gamma+t)-g(\gamma)}{t}\right)\right] \\ &=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\left( \dfrac{f( g(\gamma+t))-f( g(\gamma))}{g(\gamma+t)-g(\gamma)} \right) \right] \cdot \lim_{h\rightarrow 0}\left[ \sup_{0<|t|\le h} \left( \dfrac{g(\gamma+t)-g(\gamma)}{t}\right)\right] \\ &(?)=D^+f(c) \cdot g'(\gamma) \end{align} $$
Im not sure if the first $(?=)$ is valid but if it was true then I believe that the second $(?=)$ is true since $g$ is continuous and since $g(\gamma)=c\in(a,b)$ then for small enough $h>0$ we would have $g(\gamma +t) \in (a,b) \; \forall \; 0<|t|\le h$ because of the continuity of $g$. Then for the appropriate $h$ we would have $$\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\left( \dfrac{f( g(\gamma+t))-f( g(\gamma))}{g(\gamma+t)-g(\gamma)} \right)\right] = \lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\left( \dfrac{f( c+t)-f( c)}{t} \right)\right] $$
Is this correct?