Question

Let f(x) = \(\frac{x}{1+x}\) , x> -1. If \(g = f^{-1}\) , find the derivative of g using inverse function theorem.

Answer 1

Inverse function theorem:
Let f be differentiable and g be the inverse of f. If f'(c) \(\ne\) 0, then g is differentiable at d=f(c) and g'(d) = 1/ [f'(c)].

My attempt:
\[f'(x) = \frac{1}{(1+x)^2}\]\[g= f^{-1}\]\[g’ = \frac{1}{f'(x)} =\frac{1}{\frac{1}{(1+x)^2}} = (1+x)^2\] Somehow, I’m quite sure that I didn’t use the theorem correctly. Can anyone show me the correct way or point out my mistakes Thanks in advance!

Answer 2

It is a bit more complicated;

Answer 3

1 Express , using only f(x) = y , the variable x as a function of its image y

Answer 4

2 IN the formula u have obtained USE this expression x = some-formula(y) instead of x itself.
Then simplify - and here is ur answer only expressed in y dummy variable.

Change to x - and u have it

Answer 5

Would you mind explaining step 1, particularly ''f(x) = y''?

Answer 6

y = x/(1+x) ====>

y + xy = x ===>

x(1-y) = y

Answer 7

x= y/(1-y)

Answer 8

so plug this instead of x in YOU first answer

Answer 9

So, did you mean express x in terms of y for the first step?

Answer 10

y

Answer 11

AND plug this instead of x in YOU first answer

Answer 12

(1 + y/(1-y)) ^2

Answer 13

How does it work? I'm sorry but I don't quite understand what you're doing.

Answer 14

Have you computed the derivative of g ? - Yes you have
Buuut this must be expressed in terms of the natural variable of g.
This variable is y and NOT x

Answer 15

Sooooo ... What should a poor studiosus do if he has an answer only in terms of x ?

Answer 16

Eurica ! He should make an expression made entirely of y-dependence HWICH IS EQUIVALENT to x, and REPLACE each appearance of x with this expression ! (equal)

Answer 17

That is what I recommended - and finally applied

Answer 18

Actually, the answer given by the professor was g'(y) = 1/(1-y)^2. But he didn't explain how to get it though...

Answer 19

Last small act - the letter used in the function is in some respect pure formality ... soooo..

Answer 20

By the way, I don't understand how you find the derivative of g.... (I'm sorry again!)

Answer 21

Wait let me first check YOUR original "solution"

Answer 22

Correct

Answer 23

Huh?! Correct?!

Answer 24

Not in the sense that u should supply THAT as final expression

Answer 25

Aaaand MY ANSWER IS IDENTICAL to your proff. answer

Answer 26

Let us try to focus where're u loosing the thread - tell me

Answer 27

You did the proper thing - but only half of needed transformation

Answer 28

1. I really don't think I used that theorem correctly, but I don't know what's wrong with that.
2. I don't understand why it is in terms of y in the answer given.

Yes, I know they are identical but I'm really confused.

Answer 29

The "puenta" , essence of the answer is to use th variable y, not the variable x

Answer 30

The answer is given in y-s and NOT in x-s , is because g is formally allowed to act (compute) only upon y-s.

Answer 31

So - you must express each x in ur answer (correct, but using the wrong variable) by the expression of y(x)

Answer 32

Sorry by x(y)

Answer 33

This is what I did.

Answer 34

To give u a comparable meta-fore:

Answer 35

Suppose h(x) = x^5, then obviously h'(x) = 5*x^4

Answer 36

Now what if someone would supply u with the following instructions:

Answer 37

"LET z = x^2, and I TELL you truly, that h'(x) = 5*z^2

Answer 38

What would you notice ? And what would you correct by substitution ?

Answer 39

Firstly U would notice that his notation is a bit inconsistnet : h'9OF X) = Z-SQUARED

Answer 40

He has two different variables , ond z is NOT the right variable

Answer 41

You, would "correct" this by substituting z = x^2

Answer 42

Similar substitution was done HERE

Answer 43

I supposed \(g = f^{-1}\) is \(g(x) = f^{-1}(x)\)... Is this the mistake I made?

Answer 44

Formally speaking this is NOt a mistake.
Your ONLY mistake is not COMPLETING. The COMPLETION means using in the final expression the variable of g and not of f

Answer 45

The problem is why g the function in y not in x. I don't understand this part.

Answer 46

Why is it NOT formally a mistake you ask?
Because any variable letter can be replaced by another - they are dummy vars.

g(z). g(t), g(y) are all the same function !