# Derivative of inverse function at x=0

• feathermoon
In summary, the derivative of the inverse function at x=0 is 1/2, found by finding the derivative of the original function at x=0 and then flipping it for its inverse. The inverse function exists and f^(-1)(0)=0.
feathermoon
Derivative of inverse function at x=0 [SOLVED]

## Homework Statement

Let f(x) = x + Ln(x+1), x > -1

Find $\frac{d}{dx}$ $f^{-1}$ $|_{x=0}$; Note that f(0) = 0

## Homework Equations

$(f^{-1})'$(x) = $\frac{1}{f^{'}(f^{-1}(x))}$

(or)

$\frac{dx}{dy}$ = $1/\frac{dy}{dx}$

## The Attempt at a Solution

I attempted to find the inverse of f(x) by switching x & y and solving for y. This wouldn't work so I assume the function has no inverse.

I've read that if $f^{-1}$ is differentiable at x, the slope of the tangent line will be given by the value of the derivative of $f^{-1}$at x. Since the graph of $f^{-1}$ is obtained from the graph of f by reflection across y=x, the same is true for the tangent lines.

Then, would finding the derivative of f(x=0),

f'(x) = 1 + ($\frac{1}{x+1}$)(1)
f'(x=0) = 1 + $\frac{1}{1+0}$ = 2

and then flipping it for its inverse $\frac{1}{2}$ be a legitimate means of finding the first derivative of the inverse of f(x) at point x=0 (even if I'm not sure the inverse exists)?

Last edited:
Ok then, you only have to find f^(-1)(0). You don't need to find f^(-1)(x) for any x. What's f^(-1)(0)?

feathermoon said:

## Homework Statement

Let f(x) = x + Ln(x+1), x > -1

Find $\frac{d}{dx}$ $f^{-1}$ $|_{x=0}$; Note that f(0) = 0

## Homework Equations

$(f^{-1})'$(x) = $\frac{1}{f^{'}(f^{-1}(x))}$

(or)

$\frac{dx}{dy}$ = $1/\frac{dy}{dx}$

## The Attempt at a Solution

I attempted to find the inverse of f(x) by switching x & y and solving for y. This wouldn't work so I assume the function has no inverse.

I've read that if $f^{-1}$ is differentiable at x, the slope of the tangent line will be given by the value of the derivative of $f^{-1}$at x. Since the graph of $f^{-1}$ is obtained from the graph of f by reflection across y=x, the same is true for the tangent lines.

Then, would finding the derivative of f(x=0),

f'(x) = 1 + ($\frac{1}{x+1}$)(1)
f'(x=0) = 1 + $\frac{1}{1+0}$ = 2

and then flipping it for its inverse $\frac{1}{2}$ be a legitimate means of finding the first derivative of the inverse of f(x) at point x=0 (even if I'm not sure the inverse exists)?

Ok, you've got it. I'm pretty sure the inverse does exist. f'(x)>0 for x>(-1).

Dick said:
Ok then, you only have to find f^(-1)(0). You don't need to find f^(-1)(x) for any x. What's f^(-1)(0)?

If you mean I just find

f(x=0) = 0 + ln(1)

and take its inverse,

1/0 DNE

and I don't get anywhere.

The more I look at it I believe that the inverse just DNE and this problem is a trap.

Dick said:
Ok, you've got it. I'm pretty sure the inverse does exist. f'(x)>0 for x>(-1).

Ok, so the way I did it seems to be the only possible way to arrive at a solution? I worried about it, because I've tried the method in a test problem that I could do normally with no success:

Let f(x)= 2x^2-2, x is greater than or equal to zero.

Find d/dx f-inverse(x)|x=0 [Note than f(1)=0]

f'(x) = 4x
f'(x=0) = 0
flip for inverse 1/0

feathermoon said:
If you mean I just find

f(x=0) = 0 + ln(1)

and take its inverse,

1/0 DNE

and I don't get anywhere.

The more I look at it I believe that the inverse just DNE and this problem is a trap.

No, f^(-1)(0) is supposed to be the inverse function, not 1/f(0). f^(-1)(0)=0. You just did it correctly, why are you retracting that?

feathermoon said:
Ok, so the way I did it seems to be the only possible way to arrive at a solution? I worried about it, because I've tried the method in a test problem that I could do normally with no success:

Let f(x)= 2x^2-2, x is greater than or equal to zero.

Find d/dx f-inverse(x)|x=0 [Note than f(1)=0]

f'(x) = 4x
f'(x=0) = 0
flip for inverse 1/0

f^(-1)(0)=1. f'(1)=4. 1/f'(f^(-1)(0))=1/f'(1)=1/4. So I'm not sure what is worrying you.

Dick said:
f^(-1)(0)=1. f'(1)=4. 1/f'(f^(-1)(0))=1/f'(1)=1/4. So I'm not sure what is worrying you.

I see. I'm just having trouble finding an inverse for the original problem, and its throwing me off.

Wolfram gave me an inverse using the product log function, which I know isn't expected in this course.

Can I ask how you verified f^(-1)(0)=0? I understand everything else you've said and can finish from there.

feathermoon said:
I see. I'm just having trouble finding an inverse for the original problem, and its throwing me off.

Wolfram gave me an inverse using the product log function, which I know isn't expected in this course.

Can I ask how you verified f^(-1)(0)=0? I understand everything else you've said and can finish from there.

I'm assuming you switched back to the first problem. If f(x)=x + ln(x+1), then f(0)=0, so f^(-1)(0)=0, right?

Dick said:
I'm assuming you switched back to the first problem. If f(x)=x + ln(x+1), then f(0)=0, so f^(-1)(0)=0, right?

Yes, original problem. I see, you're finding the reflection of the point f(0)=0 across y=x (so the point (0,0)). I was making the mistake of trying to use a multiplicative inverse. I think I've got it now, thank you.

## 1. What is the derivative of the inverse function at x=0?

The derivative of the inverse function at x=0 is equal to the reciprocal of the derivative of the original function at the corresponding value of y.

## 2. How do you find the derivative of an inverse function at x=0?

To find the derivative of an inverse function at x=0, first find the derivative of the original function at the corresponding value of y. Then take the reciprocal of that value to find the derivative of the inverse function at x=0.

## 3. Can the derivative of an inverse function at x=0 be undefined?

Yes, the derivative of an inverse function at x=0 can be undefined if the derivative of the original function at the corresponding value of y is equal to 0.

## 4. How does the derivative of an inverse function at x=0 relate to the graph of the function?

The derivative of an inverse function at x=0 is equal to the slope of the tangent line to the original function at the corresponding point on the graph. This tangent line is perpendicular to the original function at that point.

## 5. Why is the derivative of an inverse function at x=0 important?

The derivative of an inverse function at x=0 is important because it allows us to find the slope of the tangent line to the original function at a specific point on the graph. This can help us understand the behavior of the function and make predictions about its values.

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