Derivative of inverse function at x=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)?

 

[Dick]

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)?

 

[Dick]

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).

 

[feathermoon]

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.

 

[feathermoon]

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]

Answer: d/dx f-inverse(0)= 1/4

f'(x) = 4x
f'(x=0) = 0
flip for inverse 1/0
(instead of 1/4 as I know the answer to be)

 

[Dick]

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?

 

[Dick]

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]

Answer: d/dx f-inverse(0)= 1/4

f'(x) = 4x
f'(x=0) = 0
flip for inverse 1/0
(instead of 1/4 as I know the answer to be)

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.

 

[feathermoon]

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.

 

[Dick]

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?

 

[feathermoon]

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.