Derivatives of Inverse Functions made simple

Derivatives of Inverse Functions can be understood easily when we first learn about inverse functions. Consider an example, f(x) = 2x and g(x)=x/2  here, the first function doubles the input values and the second function halves the input values, they are going in opposite direction. These functions f and g are called inverse functions. The inverse is showed by using ‘-1’ in the power. It does not mean x-1=1/x as in exponential functions, it just denotes that the function is the inverse of the original given function.

An Inverse Function Solver helps us to find the inverse function of the given function. We can find the inverse function y in terms of x by first solving for x and then interchanging the x and y.  The function we get by doing so will be the inverse function of the original given function. We can also find inverse function solver online.  For better understanding let us consider Inverse Function Examples as given:

Example: Find the inverse of y= 2x-3
Solution: f(x) = y = 2x-3 [original function]
Solving for x, we get
2x = y+3
x = (y+3)/2
to  get the inverse function we interchange the x, y in the above step
here, y = y-1 =(x+3)/2  is the required inverse function of 2x-3

While Graphing Inverse Functions we use the same method as we use in graphing functions. The only difference being, first we find the inverse of the given function which will be taken as y and then various x values are plugged in, to get the coordinates(x,y). These coordinates are plotted on the graph which when joined give the required graph of inverse function.
Now that we have learnt about inverse functions, let us learn about Derivative of Inverse Function. If f(x) and g(x) are inverse functions then, the derivative of inverse function is given by the formula:
g’(x) = 1/f’[g(x)]

To find the derivative of inverse function of f(x) = 2x-1, first we need to find the inverse function of f(x), which is, f-1(x)=(x+1)/2. In the next step we find the derivative of this inverse function. [f-1(x)]’= d/dx[x/2 +1/2]=1/2 [applying the derivative rule] . So, the derivative of inverse function of f(x)=2x-1 is 1/2

Let us now go through Inverse Sine Function which is given as, y= sin-1(x) which implies sin(y)=x where y lies between – pi/2 and pi/2. Evaluating an inverse sine function is same as asking what angle we need to plug into the sine function to get the value x. Let us evaluate sin-1[1/sqrt(2)]. Here we are asked for what angle ‘ y’ we arrive to the value 1/sqrt(2), written as sin(y) = 1/sqrt(2). We have already learnt that sin(45) gives us the value 1/sqrt(2) and hence we get, y = pi/4.
Learn more about how to solve Calculus Problems.