# Derivative of y=(1+1/x)^x

1. Nov 30, 2006

### haiha

How can can prove that the derivative of y=(1+1/x)^x is always positive?
Thank you

2. Nov 30, 2006

### HeilPhysicsPhysics

y=(1+1/x)^x
Let u=1+1/x,y=u^x
y=u^x
ln y=x ln u
y'/y=x/u
dy/du=[x/(1+1/x)](1+1/x)^x
=x(1+1/x)^(x-1)
du/dx=-1/x^2
dy/dx=(dy/du)(du/dx)
=x(1+1/x)^(x-1) (-1/x^2)
=-(1+1/x)^(x-1) /x

3. Dec 1, 2006

### haiha

y=(1+1/x)^x
Let u=1+1/x,y=u^x
y=u^x
ln y=x ln u
Then how come y'/y=x/u ?
I think the derivative of y will be calculated as follow:
y'/y=(x ln u)' = x*(1/u)u' + ln u (derivative of a product)
==> y' = y[x/(1+1/x)* (-1/x^2) +ln (1+1/x)]
y' = y [ln(1+1/x)-1/(x+1)]
y' = (1+1/x)^x [ln(1+1/x)-1/(x+1)]

The problem here is that I can not prove y' always positive

4. Dec 1, 2006

Your derivative is correct. Now, you should look at the domain and image of this function.

5. Dec 1, 2006

### kesh

it's not. at x = 0 it's undefined, for instance

6. Dec 1, 2006

### haiha

Yes, I should avoid the value x=0. Except from that, let say, for all x >0, why the derivative is always positive?

7. Dec 1, 2006

So? Graph the function and it will be pretty obvious that y'(x) > 0, for every x from $$(0, +\infty)$$.

8. Dec 1, 2006

### kesh

because the question asked about the function without any constraints on x's domain.

part of the problem people have while tackling questions like this is they don't emphasise the domain, as you pointed out yourself

also a graph may make it "pretty obvious", but this isn't a proof

9. Dec 1, 2006

Well, constraints on domains are self understood. It wouldn't make any sence to talk about values of the function on an interval where it isn't defined, would it?

10. Dec 1, 2006

### kesh

depends how far into analysis you get. undefined points are crucial and interesting at some levels

11. Dec 1, 2006

### haiha

Thank you all for discussing. I have graphed the curve and the derivative will leads to plus zero (+0) when x --> infinity. So in the domain of 0<x<inf, the derivative must be positive. But infact that is not a proof.

12. Dec 1, 2006

### kesh

have you tried the binomial theorem?

edit: done some calculating and that doesn't seem to help. i only have pencil and paper here, lol

Last edited: Dec 1, 2006
13. Dec 1, 2006

### HallsofIvy

Staff Emeritus
The domain of this function is $(-\infty, -1) U (0, \infty)$, not just x> 0.

14. Dec 1, 2006

### kesh

being really pedantic: the domain is any set you (or the questioner) chooses so long as the rule is defined on that set

a function is defined by the rule and the domain

i still haven't found a neat way to do proof though

15. Dec 2, 2006

### haiha

y' = (1+1/x)^x [ln(1+1/x)-1/(x+1)]
We make it : y' = A*B
Where A= (1+1/x)^x and B=ln(1+1/x/)-1/(x+1)

Because A>0 (every x>0), so we check the state of B.
The derivative of B: B'= [ln(1+1/x)-1/(x+1)]'=...=1/(x+1)[1/(x+1)-1/x] <0 with all the values of x>0. So B is a contravariant function.
We also have the fact that when x--> inf the B leads to zero, so all the values of B in the domain x>0 must be positive.
Then y' = A*B also is positive.

16. Dec 2, 2006

### Gib Z

Anyone noticed the limit of this function as it approaches infinity is e? Just some interesting info for you all..

17. Dec 4, 2006

### haiha

Yes, and the function y=(1+a/x)^x will approaches e^a when x goes to infinity. Can you prove that?

18. Dec 5, 2006

### Gib Z

There is no need to prove that, it is true by definition. The challenge would be to prove different definitions of $e$ are equivalent.

19. Dec 5, 2006

### HallsofIvy

Staff Emeritus
It is true by one definition of e. If, for example, you define
$$ln(x)= \int_1^x \frac{1}{t}dt$$
and then define exp(x) to be the inverse function to ln(x), you would need to prove that
$$\lim_{x\rightarrow \infty}\left(1+ \frac{a}{x}\right)^x= exp(a)$$

20. Dec 5, 2006

### Werg22

To simplify the question, prove that (1+1/x)^x < 0 for x < 0 and x =/= 0 and x=/= -1. For x < -1, (1+1/x)^x takes a positive value. If we define ln -|x| = - ln |x|, then the derivative is negative for x < -1. It is not always positive.