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.