Why does the proof for d/dx lnx use Euler's constant?

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In summary, the conversation discusses the use of Euler's constant in the proof of the equation d/dx ln(x) = 1/x. It is noted that the use of lim h→0 in the proof is equivalent to lim u→∞, which is why Euler's constant is relevant. However, caution should be taken when making substitutions to ensure the correct limit is used.
  • #1
aclark609
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My question has to do with Euler's constant towards the end of the proof:

d/dx ln/x = lim h → 0 1/h ln(x+h) - lnx
= lim h→0 1/h ln[(x+h/x]
= lim h→0 1/h ln(1 + h/x)
= lim h→0 ln(1 + h/x)^(1/h)(1/x)(x/1)
= lim h→0 1/x ln(1+h/x)^(x/h) ;

let u = x/h

= 1/x lim h→0 ln(1+1/u)^u
= 1/x lne
= 1/x

Doesn't Euler's constant deal with the lim h→∞ instead of 0? Are they interchangeable? If so, why?
 
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  • #2
aclark609 said:
My question has to do with Euler's constant towards the end of the proof:

d/dx ln/x = lim h → 0 1/h ln(x+h) - lnx
= lim h→0 1/h ln[(x+h/x]
= lim h→0 1/h ln(1 + h/x)
= lim h→0 ln(1 + h/x)^(1/h)(1/x)(x/1)
= lim h→0 1/x ln(1+h/x)^(x/h) ;

let u = x/h

= 1/x lim h→0 ln(1+1/u)^u
= 1/x lne
= 1/x

Doesn't Euler's constant deal with the lim h→∞ instead of 0? Are they interchangeable? If so, why?

If you let u→∞, you'll get e (which can be called Euler's number, Euler's constant usually means another number γ). Clearly, h→0 implies u→∞.
 
  • #3
EDIT: lim h-->0 should be lim u-->0
 
Last edited:
  • #4
aclark609 said:
EDIT: lim h-->0 should be lim u-->0

Why? When you make a substitution or a change of variables, you have to check how the new variable (i.e. u) depends on the original one (i.e. h). Otherwise, you'd get ridiculous results like the following (let y=2x)
[tex]2=\lim_{y\rightarrow 2} y = \lim_{x\rightarrow 2} 2x=4[/tex]

EDIT: My point is that your final limit should be [itex]u\rightarrow \infty[/itex], not [itex]u\rightarrow 0[/itex].
 
Last edited:
  • #5
Aaaahhh. You're right. However if I allow h/x = u, then then lim u -->0 would be correct.
 

What is the definition of lnx?

The natural logarithm of x, denoted as ln(x), is the inverse function of the exponential function. It is the power to which e (the base of natural logarithms) must be raised to equal x.

What is the derivative of lnx?

The derivative of lnx is 1/x. This means that when you take the derivative of ln(x), you will get 1/x as the result.

How do you prove that d/dx lnx = 1/x?

To prove that d/dx lnx = 1/x, we can use the definition of the derivative. We start by writing lnx as a power of e, using the fact that ln(x) = e^y when x = e^y. Then, we can use the power rule for derivatives and simplify to get 1/x as the final result.

What is the intuition behind the derivative of lnx being 1/x?

The derivative of lnx being 1/x means that as x changes, ln(x) changes at a rate of 1/x. This can be interpreted as the slope of the tangent line to the curve at any point on the graph of ln(x), which is always 1/x. This intuitive understanding is based on the definition of the derivative as the rate of change of a function.

Can the derivative of lnx be extended to other logarithmic functions?

Yes, the derivative of lnx can be extended to other logarithmic functions using the same principles. For example, the derivative of log2(x) would be 1/(xln2). This can be shown using the same steps as proving the derivative of lnx, but using the properties of logarithms for different bases.

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