Left hand derivative and Right hand derivative of the inverse of a non-differentiable function Let $f(x)$ be an injective function with domain $[a,b]$ and range $[c,d]$. 
If $\alpha$ is a point in $(a,b)$ such that $f$ has left hand derivative $l$ and right hand derivative $r$ at $x=\alpha$ 
with both $l$ and $r$ being non-zero different and negative ,
then prove that the left hand derivative and the right hand derivative of $f^{-1}(x)$ at $x=f(\alpha)$ are $\frac{1}{r}$ and $\frac{1}{l}$ respectively.
My Attempt:
If $g(x)$ be the inverse of $f((x)$ then $f(g(x))=x$ which gives $g'(x)=\frac{1}{f'(g(x))}$.
So if $x=f(\alpha)$ then $g'(f(\alpha))=\frac{1}{f'(g(f(\alpha)))}=\frac{1}{f'(\alpha)}$.
But now how do I get the left hand derivative and the right hand derivative.
I could frame an example
 $f(x)=\left\{
\begin{array}{ll}
      \frac{1}{x} & 0<x<2 \\      
      \frac{2}{x^2} & x\geq 2 \\
      \end{array} 
\right.$ 
Left hand derivative at $x=2$ equals $\frac{-1}{4}=l$
Right hand derivative at $x=2$ equals $\frac{-1}{2}=r$
$f^{-1}(x)=\left\{
\begin{array}{ll}
      \\\sqrt {\frac{2}{x}} & 0<x\leq \frac{1}{2} \\      
      \frac{1}{x} & x> \frac{1}{2} \\
      \end{array} 
\right.$ 
Left hand derivative of $f^{-1}(x)$ at $x=\frac{1}{2}$ equals $-2=\frac{1}{r}$
Right hand derivative of $f^{-1}(x)$ at $x=\frac{1}{2}$ equals $-4=\frac{1}{l}$
But I am not able to get a proper method.
 A: We have $f'_-(\alpha)=l < 0$ and $f'_+(\alpha)=r < 0$, with $l \neq r$. That implies, for $h > 0$ small enough, that $f(\alpha - h) > f(\alpha) > f(\alpha + h)$.
Since $f$ has a left and a right derivative at $\alpha$, it is continuous at $\alpha$. 
Let $\beta = f(\alpha)$ and given $h > 0$, let $k_1 = f(\alpha + h) - f(\alpha)$ and $k_2 = f(\alpha - h) - f(\alpha)$. Note that $k_1 < 0$ and $k_2 > 0$.
Then $\frac{g(\beta + k_1) - g(\beta)}{k_1}=\frac{h}{f(\alpha + h) - f(\alpha)} \to_{h \to 0^+} \frac{1}{r}$,
and $\frac{g(\beta + k_2) - g(\beta)}{k_2}=\frac{-h}{f(\alpha - h) - f(\alpha)} \to_{h \to 0^+} \frac{1}{l}$.
A: Let $\beta=f(\alpha)$..
First, since $f$ is differentiable at $x=\alpha$ then $f$ is continuous at $x=\alpha$ and consequently 
$$x\rightarrow \alpha^{+}\Longrightarrow f(x)\rightarrow f(\alpha)=\beta.$$
Since $f^{\prime+}(\alpha)=\lim_{x\rightarrow \alpha^{+}}{\frac{f(x)-f(\alpha)}{x-\alpha}}=r$ then 
$$\frac{1}{{(f^{-1})}^{\prime}(\beta)}=\lim_{y\rightarrow \beta}{\frac{y-\beta}{f^{-1}(y)-f^{-1}(\beta)}}=r$$.
Observe that we denoted $y=f(x)$, $x\in [a,b]$ or equivalently $x=f^{-1}(y)$, $y\in [c,d]$. More importantly, we have that $f^{-1}(y)-f^{-1}(\beta)\neq 0$ since $f$ is injective and $x\neq \alpha$.
The left derivative of the inverse can be obtained analogously.
A: 
Picking up from both answers given I was able to understand and consequently plotted the graph depicting $y=f(x)$ and $y=f^{-1}(x)$
