Derivative of the Inverse function

Let y=f(x)  and x=g(y)  are inverse functions of each other. Then

x=g(y)=g(f(x))  and y=f(x)=f(g(y))

Considering x=g(y)=g(f(x)) and differentiating both sides with respect to x , we get

1= g (f(x)) f (x)

or, 1= g (y) f (x)

Similarly, considering y=f(x)=f(g(y)) and differentiating both sides with respect to y , we get

1= f (g(y)) g (y)

or, 1= f (x) g (y)

So, it does not matter whether we consider g(f(x)) or, f(g(y)) , we arrive at the same relation between the derivatives of the original function y=f(x) and the inverse function g(y)

1= f (x) g (y)

or, g (y)= 1 f (x)

If we write dx dy for g (y) and dy dx for f (x) then we have

dx dy = 1 dy dx

The usual notation for the inverse of the function y=f(x) is given by g(y)= f 1 (y)

So, what we have is ( f 1 ) (y)= 1 f (x)

What this tells us is that if we know the value of the derivative of the function f(x) at x=a , then we can easily calculate the derivative of its inverse function f 1 (y) at the point y=b=f(a) as follows:

( f 1 ) (b)= 1 f (a)

Consider the function f(x)= x 3 +4x , which has an inverse function f 1 (x) . It is not easy to find the exact definition of this inverse function. Still, without knowing the exact definition of this inverse function, we can determine the value of ( f 1 ) (5) .

Since f(1)=5 , we have ( f 1 ) (5)= 1 f (1) = 1 7 . We have used the fact that f (x)=3 x 2 +4 and f (1)=7 .

While we have to be careful in what we say, but we can think of it this way. The derivative of the inverse is the inverse of the derivative. :-)

Also, think of it this way. Suppose A is moving 4 times faster than B. You can construct a function to describe this. Now think of the function that describes how fast is B moving with respect to A. Clearly, this function is the inverse of the original function. And, you can say B is moving 1 4 times faster than A. :-)