Derivative of inverse functions Show graphically that
$$
(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}. \\
$$
PS: I know that $f(x)$ and $f^{-1}(x)$ are symmetrical about the line $y=x$. It should be proved  by showing that multiplying the slopes of the tangents of points $x$ and $f(x)$ equals 1.

What does mean "graphically"? It is not clear to me. 
We know that $f(f^{-1}(x))=x$ by definition
Thus $[f(f^{-1}(x))]'=f'(f^{-1}(x))(f^{-1})'(x)=1$
And then $(f^{-1})'(x)=\dfrac{1}{f'(f^{-1}(x))}$
But this is not clearly "graphical"...
