Why is this?

Reuel · 2011-04-18 08:39:08

Why does the derivative of f set equal to the reciprocal of its inverse return the original function when solved for f?

For example, consider a generic quadratic such as

the inverse of which is the quadratic formula (with y intact)

considering only the positive form of the solution.

If we differentiate f we get

and if we differentiate f's inverse, g(y) we get

If we set the derivative of f equal to the reciprocal of g, its inverse, we have

and if we solve this equality for y we get

which, of course, is our original function.

I showed this to my math teacher and he didn't really answer my question. I am curious as to why this is true. I know that the derivative of f will have reciprocal derivatives in the derivative of its own inverse, but why does setting the derivative of f equal to the reciprocal of its inverse result in the original function? Is there a proof or anything?

Thanks.

Last edited by Reuel (2011-04-18 08:40:28)

gAr · 2011-04-18 16:57:36

Hi Reuel,

Use chain rule.

Reuel · 2011-04-19 12:41:33

Unless I am mistaken, which I may very well be, that seems to be a proof of how the derivative of the inverse will be a reciprocal of the derivative of the function.

But why is setting the derivative of the function equal to the derivative of the inverse of the function and solving for y going to result in the original function?

I don't get it.

gAr · 2011-04-19 16:09:54

Hmmm, using y, f(x), g(y) are confusing.

Reuel · 2011-04-20 09:48:18

I am not a master of the theory behind calculus. Are differential operators a thing that can be canceled out the same way a variable may be canceled?

If we want to set the derivative of f equal to the reciprocal of the derivative of the inverse of f, we might write

where f-inverse is a function x in terms of y. That is,

Now I know that in differential equations one can move the dx over to the other side the equation when separating the variables. Does that mean that other operations can be used as well? Such as cancellation?

If we cross-multiple we can get

and if d is a thing that may be canceled out, we could then get

and, if so, since

we can reduce to

That may be a complete perversion of mathematics. Sorry, if so.

Does what I just did make sense? Is that one way of showing how the function y can be taken from the two derivatives without calculus?

Thanks so much for the help.

gAr · 2011-04-20 18:01:51

You cannot cancel the "d" like that.
E.g.

What you have done in your first post is explained in #4

Reuel · 2011-04-20 22:18:28

Yes, if y is f(x) or y(x) then its inverse would be x(y) or just x. The inverse has to be in terms of y so there is a y to solve for.

I am sorry to be confusing but that much, at least, is correct if not with clarity.

I see that canceling d can sometimes be a bad idea.

As I said, the inverse need be in terms of y so there is a y to solve for. I still do not get why solving for that y produces the function and not something else when we are starting with derivatives. I guess I will just keep looking at it.

123ronnie321 · 2011-04-20 22:22:18

Hi Reuel,

I am sure you will understand it if you draw some function and its inverse like the graphs shown on this page.
I hope it helps.

http://www.understandingcalculus.com/chapters/13/13-1.php

gAr · 2011-04-21 02:17:49

Hmmm, you are simply going a circle.

Your g(y) is actually x, and f(x) is y. So you must arrive at the original function when you solve for y, since

I think you should go through some derivatives of inverse functions
E.g.
http://www.analyzemath.com/calculus/Differentiation/derivative_inverse.html

Math Is Fun Forum

#1 2011-04-18 08:39:08

Why is this?

#2 2011-04-18 16:57:36

Re: Why is this?

#3 2011-04-19 12:41:33

Re: Why is this?

#4 2011-04-19 16:09:54

Re: Why is this?

#5 2011-04-20 09:48:18

Re: Why is this?

#6 2011-04-20 18:01:51

Re: Why is this?

#7 2011-04-20 22:18:28

Re: Why is this?

#8 2011-04-20 22:22:18

Re: Why is this?

#9 2011-04-21 02:17:49

Re: Why is this?

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