Drawn below is the graph of
f (x) = x2 on the interval
(0,∞), and its inverse on
that interval,
f-1(x) = 
. Also drawn on the graph are the tangents to the graph of
f (x) at (2,4), and the
tangent to the graph of
f-1(x) at the reflected point (4,2).
In the case above,
f'(x) = 2x and
(f-1)'(x) = 
It seems that at least in this case, the derivative of
f at
(a, b) is the reciprocal of the
derivative of
f-1 at
(b, a). This in fact holds true in all cases. In general, it can be
said that if
(a, b) is a point on
f then
(b, a) is a point on
f-1, and
(f-1)'(b) = 
.
To make this statement even more applicable, we should now try to find a formula for
(f-1)'(x). From the formula above, if we let b = x, then a = f-1(x), so that the
following more general statement may be written: