# Inverse function of a quarter circle gives me same function

1. Aug 11, 2015

### jaysquestions

Is this normal? it doesn't seem correct.
The equation for the portion of circle with radius 1 unit in the 1st Quadrant is:
$y = f(x) = \sqrt{1-x^2}$ Domain is 0<x

But when I calculate f'(x) I also get
$f'(x) = \sqrt{1-x^2}$

2. Aug 11, 2015

### JonnyG

You are calculating the derivative incorrectly.

3. Aug 11, 2015

### Staff: Mentor

Please show the work to get the above (which is wrong).
What does your work have to do with inverse functions? There is a formula for the derivative of the inverse of a function. Is that what you're doing?

4. Aug 11, 2015

### willem2

If you draw graph of the function and the line y=x you should quickly see that reflection about y=x doesn't change the graph.

Assuming you meant f'(x) to be the inverse function of f(x) and not the derivative.

5. Aug 11, 2015

### jaysquestions

Sorry for the confusion I didn't mean derivative at all I meant inverse. I was having probs trying to make latex show a superscript -1. (actually I still cant do it)

6. Aug 11, 2015

### jaysquestions

I meant inverse, i thought it was the same as the derivative symbol, plus I couldn't get f^-1(x) in latex

7. Aug 11, 2015

### Staff: Mentor

$f^{-1}(x)$

Renders as $f^{-1}(x)$

8. Aug 12, 2015

### HOI

Yes, inverse functions always "reflect about y= x". And the function, $y= \sqrt{1- x^2}$ is the same as $x^2+ y^2= 1$ in the first quadrant it is symmetric in x and y- solving for x or y give exactly the same thing- the function is the same as its inverse function. There a number or functions that are the same as their inverses:
y= x, y= 1- x, y= 1/x, ...

9. Aug 12, 2015

### jaysquestions

thanks for the answers and the latex tip for inverse functions.
Do inverse functions have some sort of relationship to derivatives and if so what is the relationship? (just in generality I mean) I know that derivative of fcn is the slope and that inverse of fcn reflects about x-axis, so do they all relate together in some way?
thanks

10. Aug 18, 2015

### HallsofIvy

If $f(x)$ has inverse function $f^{-1}(x)$ then $f^{-1}(f(x))= 1$. Differentiating both sides of that with respect to x, using the "chain rule" on the left, $$\frac{df^{-1}(f(x))}{dx}\frac{df}{dx}= 1$$ so $$\frac{df^{-1}(f(x))}{dx}= \frac{1}{\frac{df}{dx}}$$