Inverse function of a quarter circle gives me same function

[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
$f'(x) = \sqrt{1-x^2}$
I thought inverse functions always reflect about y=x. Please help...confused...thanks

 

[JonnyG]

You are calculating the derivative incorrectly.

 

[Mark44]

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
$f'(x) = \sqrt{1-x^2}$

Please show the work to get the above (which is wrong).

I thought inverse functions always reflect about y=x. Please help...confused...thanks

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?

 

[willem2]

But when I calculate f'(x) I also get
$f'(x) = \sqrt{1-x^2}$
I thought inverse functions always reflect about y=x. Please help...confused...thanks

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.

 

[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 can't do it)

 

[jaysquestions]

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?

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

 

[Mark44]

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

##f^{-1}(x)##

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

 

[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, ...

 

[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

 

[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 $$\frac{df^{-1}(f(x))}{dx}= \frac{1}{\frac{df}{dx}}$$