Is the Derivative of (1+1/x)^x Always Positive?

  • Thread starter haiha
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In summary, the given function y=(1+1/x)^x has a derivative that is always positive for all values of x greater than 0. However, the domain of this function is actually (-∞, -1) U (0, ∞), not just x > 0. For x < -1, the derivative is negative.
  • #1
haiha
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How can can prove that the derivative of y=(1+1/x)^x is always positive?
Thank you
 
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  • #2
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
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
haiha said:
y' = (1+1/x)^x [ln(1+1/x)-1/(x+1)]

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

Your derivative is correct. Now, you should look at the domain and image of this function.
 
  • #5
haiha said:
How can can prove that the derivative of y=(1+1/x)^x is always positive?
Thank you
it's not. at x = 0 it's undefined, for instance
 
  • #6
Yes, I should avoid the value x=0. Except from that, let say, for all x >0, why the derivative is always positive?
 
  • #7
kesh said:
it's not. at x = 0 it's undefined, for instance

So? Graph the function and it will be pretty obvious that y'(x) > 0, for every x from [tex](0, +\infty)[/tex].
 
  • #8
radou said:
So?
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
kesh said:
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

Well, constraints on domains are self understood. It wouldn't make any sense to talk about values of the function on an interval where it isn't defined, would it?
 
  • #10
radou said:
Well, constraints on domains are self understood. It wouldn't make any sense to talk about values of the function on an interval where it isn't defined, would it?
depends how far into analysis you get. undefined points are crucial and interesting at some levels
 
  • #11
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
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:
  • #13
The domain of this function is [itex](-\infty, -1) U (0, \infty)[/itex], not just x> 0.
 
  • #14
HallsofIvy said:
The domain of this function is [itex](-\infty, -1) U (0, \infty)[/itex], not just x> 0.
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
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
Anyone noticed the limit of this function as it approaches infinity is e? Just some interesting info for you all..
 
  • #17
Yes, and the function y=(1+a/x)^x will approaches e^a when x goes to infinity. Can you prove that?
 
  • #18
There is no need to prove that, it is true by definition. The challenge would be to prove different definitions of [itex]e[/itex] are equivalent.
 
  • #19
It is true by one definition of e. If, for example, you define
[tex]ln(x)= \int_1^x \frac{1}{t}dt[/tex]
and then define exp(x) to be the inverse function to ln(x), you would need to prove that
[tex]\lim_{x\rightarrow \infty}\left(1+ \frac{a}{x}\right)^x= exp(a)[/tex]
 
  • #20
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.
 

1. What is the formula for the derivative of y=(1+1/x)^x?

The formula for the derivative of y=(1+1/x)^x is y' = (1+1/x)^x * ln(1+1/x) * (1/x^2) - (1+1/x)^(x-1).

2. How do you simplify the derivative of y=(1+1/x)^x?

To simplify the derivative of y=(1+1/x)^x, first rewrite the equation as y=e^(x*ln(1+1/x)). Then, apply the chain rule by multiplying the derivative of the exponent (which is ln(1+1/x)) by the derivative of the inner function (which is x). Finally, use the power rule to find the derivative of e^(x*ln(1+1/x)).

3. What is the domain of the function y=(1+1/x)^x?

The domain of y=(1+1/x)^x is all real numbers except for x=0.

4. How does the graph of y=(1+1/x)^x look like?

The graph of y=(1+1/x)^x is a smooth curve that approaches the x-axis as x approaches infinity. It has a minimum point at x=e-1 and a vertical asymptote at x=0.

5. Can the derivative of y=(1+1/x)^x be negative?

Yes, the derivative of y=(1+1/x)^x can be negative. It is negative when x is between 0 and e-1. This means that the slope of the graph is decreasing at those points.

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