Let $f:[a,b]\to\Bbb R$ and show that for any point $c\in[a,b]$ we have $D^+(f(x),c)=-D_-(f(x),c)$. Here $D^+$ and $D_-$ indicate Dini derivatives: $$D^+(f(x),c)=\limsup_{h\to 0^+}\frac{f(c+h)-f(c)}{h}$$ $$D_-(f(x),c)=\liminf_{h\to 0^-}\frac{f(c+h)-f(c)}{h}$$
It seems like this is related to a change of variables. If we write
$$ D^+(f(x),c)=\lim_{h\to 0^+} \sup_{0< t\le h}\frac{f(c+t)-f(c)}{t} $$
and set $i=-h$ then
$$ D^+(f(x),c)=\lim_{i\to 0^-}\sup_{i \le t < 0}\frac{f(c-t)-f(c)}{-t} = -\lim_{i\to 0^-}\inf_{i\le t<0}\frac{f(c-t)-f(c)}{t}$$
Now the lim and inf look like they need to for $D_-$ but the stuff on the inside looks wrong for the definition of the derivative. I know when the limit isn't handed that doesn't matter, but here it seems like it should matter--I should be able to simply change this to $\frac{f(c+t)-f(c)}{t}$.
I also have from a professor the following proof, which claims to establish the equality.
Let $g(x)=f(-x)$.
$$ D^+(g(x),c)=\limsup_{h\to 0^+}\frac{g(c+h)-g(c)}{h} $$ $$= \limsup_{h\to 0^+} \frac{f(-(c+h))-f(-c)}{h} $$ $$=-\liminf_{h\to 0^+}\frac{f(-c)-f(-c-h)}{h} =-D_-(f(-x),c)$$
However, as I try to prove the last equality, that seems to run into the same issue as above.
To be explicit,
\begin{align*} &\quad \varliminf_{h \to 0^-} \frac {f(c + h) - f(c)}h \\ &= \lim_{h \to 0^-} \inf_{h \leqslant A \leqslant 0} \frac {f(c + A) - f(c)}A\\ &= \lim_{-h \to 0^+} \inf_{-h \geqslant -A \geqslant 0} -\frac {f(c - (-A)) - f(c)}{-A}\\ &= \lim_{k \to 0^+} \inf_{k \geqslant B \geqslant 0} \frac {f(c) - f(c-B)}{B} \tag {$k = -h, B = -A$}\\ &=\varliminf_{h \to 0^+} \frac {f(c) - f(c-h)}{h} \end{align*}