# Inverse and original function relationships

1. Jan 8, 2014

### MathewsMD

Just curious:

Are there any unique relationships b/w the inverse of a function and the original, specifically when considering the derivative and integral?

2. Jan 8, 2014

### DrewD

wiki for derivative and Wolfram for the integral. Is that what you're looking for?

3. Jan 9, 2014

### HallsofIvy

Use the chain rule on $f(f^{-1}(x))$ to get the first and "integration by parts" to get the second.

4. Jan 9, 2014

### MathewsMD

Just wondering...if you have an original function $f(x) = y$ and the inverse, $g(y) = x = f^{-1} (f(x))$.

Then differentiating $g(y)$ wrt to x you get:

$\frac {d(g(y)}{dy} \frac {dy}{dx} = g'(y)y' = g'(f(x))f'(x) = f'^{-1} (f(x)) f'(x)$

I'm just slightly confused on how the article derived $f'^{-1} (x) = \frac {1}{f' ( f^{-1} (x))}$ since all it says is that chain rule is used, specifically here: http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation#Additional_properties.

Also, just a little lower in the article, it states:

$\frac {d^2y}{dx^2} . \frac {dx}{dy} + \frac {d^2x}{dy^2} . (\frac {dy}{dx})^2 = 0$

All I can really simplify this to by simple manipulation is:

$\frac {(d^2y)(dy^2)}{(d^2x)(dx^2)} = -(\frac {dy}{dx})^3$

This isn't really helpful and I don't exactly know what else to do from here. Any insight on taking a different approach or how to move on would be greatly appreciated. I feel like I'm just not applying chain rule correctly, but any help would be great!

Last edited: Jan 9, 2014
5. Jan 9, 2014

### DrewD

Just use $g(f(x))=x$ and differentiate. I think it actually works faster if you use $f(g(x))=x$ like HoI, but both work fine.

6. Jan 10, 2014

### MathewsMD

Hmm....I must say that my understanding of Liebniz notation is not too strong.

For example, what exactly does this represent (if $y = f(x)$):

$\frac {d^3y}{dx^3}$

Is it simply $\frac {1}{f'''^{-1}(x)}$ ?

Also, is $\frac {d^2y}{dx^2} . \frac {dx}{dy} + \frac {d^2x}{dy^2} . (\frac {dy}{dx})^3 = 0$

equal to $\frac {1}{f''^{-1}(x)}. \frac {1}{f^{-1}} + \frac {1}{f''(x)}. (f'(x))^3 = 0$

Besides rewriting it, I am still a little confused on how this was simplified in the next steps in the link.

Last edited: Jan 10, 2014
7. Jan 10, 2014

### MathewsMD

I am also going through the expression:

$f^{-1}(x) = \int \frac {1}{f'(f^{-1}(x))}dx + c$

By taking the derivative of both sides, what I simplify this to is:

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

I just don't exactly see how this result was derived using Chain Rule in the link.

8. Jan 15, 2014