# Derivative of an inverse function

• JanClaesen
In summary: Instead, you need to find g by using the fact that f(g(y)) = y. Therefore, you can solve for g(y) and then differentiate to find g'(y).In summary, we discussed finding the inverse of a function f(x) and how to properly use the notation for inverse functions. It was clarified that g'(y) = 1/f'(x) and integration tricks cannot be used to find the inverse. Instead, we must use the fact that f(g(y)) = y and solve for g(y) to find the inverse function.
JanClaesen
f(x) = cosh^2(x)+sinh(2x) = y
f'(x) = sinh(2x)+2cosh(2x) = 3e^(2x) + e^(-x) = y'

Let g(y) be the inverse of f(x):
g'(y) = 1 / f'(x) = 1 / [3e^(2y) + e^(-2y)] = e^(2y) / [4e^(2y) + 1]

Integrating gives: [ 3^(1/2)/3 ]*arctan[ 3^(1/2) * e^(2y) ] + C

Now when I plotted this function it looked in no way like the inverse of f(x), so where have I gone wrong?

Thank you

JanClaesen said:
Let g(y) be the inverse of f(x):
g'(y) = 1 / f'(x) = 1 / [3e^(2y) + e^(-2y)] = e^(2y) / [4e^(2y) + 1]

you just replaced y by x. What you have to do is use the relation y = g(x), where g is the inverse of f. Given your definition of f it might be hard to find an expression for this inverse

Hello
I don't understand why g(y) isn't the inverse?
f: x -> y and g: y -> x
I still don't understand why integrating 1/f'(g(y)) doesn't yield me g(y)

JanClaesen said:
Let g(y) be the inverse of f(x):
g'(y) = 1 / f'(x) = 1 / [3e^(2y) + e^(-2y)] = e^(2y) / [4e^(2y) + 1]
This is incorrect. Although you write g'(y)=1/f'(x), you are just using g'(y)=1/f'(y).

As said above, you should be using g'(y)=1/f'(x) where x and y are such that y=f(x)! The relation between x and y is crucial.

JanClaesen said:
g'(y) = 1 / f'(x) = 1 / [3e^(2y) + e^(-2y)] = e^(2y) / [4e^(2y) + 1]

To clarify what was said before, how is f'(x), a function of x, expressed completely in terms of y?

You should have g'(y) = 1/[3e^(2x) + e^(-2x)] = e^(2x)/(4e^(2x) + 1)

In which case your integration tricks won't work.

JanClaesen said:
Hello
I don't understand why g(y) isn't the inverse?
f: x -> y and g: y -> x
I still don't understand why integrating 1/f'(g(y)) doesn't yield me g(y)

That would be true. However, you seem to have tried to integrate 1/f'(y) to recover g and that does not work.

## What is the derivative of an inverse function?

The derivative of an inverse function is the rate of change of the inverse function at a given point, which can be calculated using the inverse function theorem.

## How do you find the derivative of an inverse function?

To find the derivative of an inverse function, first take the derivative of the original function, then swap the x and y variables and solve for the derivative of the inverse function.

## Why is the derivative of an inverse function important?

The derivative of an inverse function is important because it allows us to calculate the slope or rate of change for inverse functions, which is crucial in many applications of mathematics and science.

## What is the difference between the derivative of an inverse function and the derivative of a regular function?

The main difference between the derivative of an inverse function and a regular function is that the derivative of an inverse function involves swapping the x and y variables, while the derivative of a regular function does not.

## Can the derivative of an inverse function be negative?

Yes, the derivative of an inverse function can be negative. This indicates that the inverse function is decreasing at that point, rather than increasing.

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