Every one-to-one function f has an inverse function f-1 which essentially reverses the operations performed by f.

More formally, if f is a one-to-one function with domain D and range R, then its inverse f-1 has domain R and range D. f-1 is related to f in the following way: If f (x) = y, then f-1(y) = x. Written another way, f-1(f (x)) = x.

Example: f (x) = 3x - 4. Find f-1(x).

The procedure for finding f-1(x) from f (x) involves first solving for x in terms of y.

y = 3x - 4  
x =  

Now switch the variables x and y in the equation to generate the inverse:

y =  
f-1(x) =  

A function and its inverse are related geometrically in that they are reflections about the line y = x:

Figure %: A function and its inverse are symmetric with respect to the line y = x

Thus, if (a, b) is a point on the graph of f, then (b, a) is a point on the graph of f-1.

The Derivative of the Inverse

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).

Figure %: The tangent lines drawn at corresponding points on the graphs of f and f-1

What is the relationship between f (x) at (a, b) and f-1(x) at (b, a)?

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:

(f-1)'(x) =    

Note that in Leibniz notation, this becomes an intuitive: