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feathermoon

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**Derivative of inverse function at x=0 [SOLVED]**

## Homework Statement

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

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

## Homework Equations

[itex](f^{-1})'[/itex](x) = [itex]\frac{1}{f^{'}(f^{-1}(x))}[/itex]

(or)

[itex]\frac{dx}{dy}[/itex] = [itex]1/\frac{dy}{dx}[/itex]

## 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 [itex]f^{-1}[/itex] is differentiable at x, the slope of the tangent line will be given by the value of the derivative of [itex]f^{-1}[/itex]at x. Since the graph of [itex]f^{-1}[/itex] 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 + ([itex]\frac{1}{x+1}[/itex])(1)

f'(x=0) = 1 + [itex]\frac{1}{1+0}[/itex] = 2

and then flipping it for its inverse [itex]\frac{1}{2}[/itex] 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)?

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