Why is ln(2) in the Derivative of 2^u?

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Discussion Overview

The discussion revolves around the derivative of the function f(u) = 2^u, specifically addressing the presence of ln(2) in the derivative expression f'(u) = (2^u) * ln(2). Participants explore the reasoning behind this inclusion, the properties of logarithms, and the implications of using different logarithmic bases.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how ln(2) appears in the derivative of 2^u, despite understanding the chain rule.
  • Another participant explains the differentiation process using logarithmic properties, showing how ln(2) arises from the differentiation of y = 2^u.
  • There is a discussion about the interchangeability of "Log" and "ln," with some participants asserting that "Log" can refer to the natural logarithm in certain contexts, while others argue that it typically refers to logarithm base 10.
  • Concerns are raised about the potential for confusion when using different logarithmic bases, particularly in graphing contexts.
  • A participant questions the treatment of constants in differentiation, specifically why the derivative of ln(2) is not taken, leading to a clarification that ln(2) is a constant.
  • Further inquiries are made about the rules of differentiation and the application of the product rule, with participants seeking to understand the implications of these rules on their calculations.

Areas of Agreement / Disagreement

Participants generally agree that ln(2) is a constant and that its derivative is zero. However, there remains disagreement and confusion regarding the interchangeability of logarithmic terms and the implications of using different bases in calculations.

Contextual Notes

Some participants express uncertainty about the rules of differentiation, particularly regarding constants and the product rule. There is also a lack of consensus on the use of "Log" versus "ln" in various mathematical contexts.

Xtensity
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i2vo1z.png


in that picture is a practice problem from a site I am using. I have highlighted the part of it that is confusing me.

In the picture, if f(u) = 2^u , then how does f ' (u) = (2^u)*ln(2) ?

I understand the chain rule aspect of the problem, but I have no clue where that "ln" came from or what rule that is. I understand ln is a natural log, which is Log Base e... though I have no idea why it was introduced into the problem as part of the deriavtive

I have been searching online for the past hour and can not figure this out.(I would be posting a question as simple as this on YahooAnswers, but YahooAnswers has stopped working for some reason =\)
 
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I am not sure if there is name for it, but here is how it worked.

y=2u⇒lny=ln2u⇒lny=uln2

Differentiating both sides w.r.t. u

(1/y)dy/du=ln2 ⇒ dy/du= yln2. But y=2u.

In the end, we have

dy/du=2uln2.
 
Is that Ln interchangeable with Log? I used this other site for "Step by Step derivatives"

http://library.wolfram.com/webMathematica/Education/WalkD.jsp

It gave me

[PLAIN]http://library.wolfram.com/Explore/MSP/MSP019c0981if4bd1ch300006adhih2g0hfi65ed?MSPStoreType=image/gif&s=2

The only difference between that and the other site is Log vs Ln, which give 2 very different answers.
 
Last edited by a moderator:
log is sometimes used to mean logarithm base 10 and sometimes used to mean natural logarithm. Basically, once you get to a certain point in math you decide that you never actually use log base 10 so you just use log to mean natural logarithm. In this case, wolfram uses log to mean natural logarithm
 
Office_Shredder said:
log is sometimes used to mean logarithm base 10 and sometimes used to mean natural logarithm. Basically, once you get to a certain point in math you decide that you never actually use log base 10 so you just use log to mean natural logarithm. In this case, wolfram uses log to mean natural logarithm

Er... would that not produce 2 totally different answers when graphed? I don't see how or why they would be used interchangeably. Log = Log base 10, and Ln = log base e?
 
Xtensity said:
Er... would that not produce 2 totally different answers when graphed? I don't see how or why they would be used interchangeably. Log = Log base 10, and Ln = log base e?

No they aren't interchangeable functions, it's just that people use the term "Log" and "ln" interchangeably. Mathematica (what wolfram alpha is based off of) uses Log to mean natural log. If you really meant base-10 log, you'd specify it.
 
Pengwuino said:
No they aren't interchangeable functions, it's just that people use the term "Log" and "ln" interchangeably. Mathematica (what wolfram alpha is based off of) uses Log to mean natural log. If you really meant base-10 log, you'd specify it.

So if I were to use it in a graphing calculator, which one would I use Log or Ln? Since the calculator does not use Log or Ln interchangeably, on the calculator the defined functions are specific.
 
Xtensity said:
So if I were to use it in a graphing calculator, which one would I use Log or Ln? Since the calculator does not use Log or Ln interchangeably, on the calculator the defined functions are specific.

The calculator would use Log as log-10. If you ever actually see both options, then they're trying to distinguish between the two.
 
Pengwuino said:
The calculator would use Log as log-10. If you ever actually see both options, then they're trying to distinguish between the two.

I know that... but I asked which one would I use for the correct derivative.

If I have 2^u and I want to plot the correct derivative, do I use (2^u)(ln(2)) or (2^u)(log(2))... because those 2 would result in 2 different graphs... and whichever one is used, why is it used over the other?
 
  • #10
The correct one is ln(2). wolfram (and many mathematicians) uses the notation 'log' to denote 'ln' because logarithm base 10 is not commonly used.

Log=log base 10 on your graphing calculator, but it's not uncommon to find people using log to be log base e. Of course this is a different function from log base 10, but it's just something you'll have to watch out for
 
  • #11
Ok I understand we use Ln... though I'm having issues with this earlier post:

rock.freak667 said:
I am not sure if there is name for it, but here is how it worked.

y=2u⇒lny=ln2u⇒lny=uln2

Differentiating both sides w.r.t. u

(1/y)dy/du=ln2 ⇒ dy/du= yln2. But y=2u.

In the end, we have

dy/du=2uln2.

What happen to the "u" in the bolded step above? I understand that d/dx ln(y) = 1/y... but on the other side, where did the u disappear to?
 
  • #12
What is d/du u?
 
  • #13
l'Hôpital said:
What is d/du u?

Is it? I only started teaching myself calculus 1-2 weeks ago so I'm not entirely sure on the rules of the d/dx operators, etc... Do they operate as fractions or? If anyone could provide a link to the rules on those operators that would be great.
 
  • #14
Allow me to rephrase the question. What is the derivative of u with respect to u?
 
  • #15
l'Hôpital said:
Allow me to rephrase the question. What is the derivative of u with respect to u?

1... but if the derivative of u was taken, then why was the derivative of ln(2) not taken? Wouldn't the derivative of ln(2) = 1/2? Is there a rule for why that derivative isn't taken?
 
  • #16
ln 2 is a constant. : )
 
  • #17
l'Hôpital said:
ln 2 is a constant. : )

So derivatives of constants aren't taken? If so, this is a rule I have never ran into :P
 
  • #18
??

They are taken, it's just that they are zero. Generally, if

<br /> f(x) = c<br />

For some constant c, then
<br /> <br /> f&#039;(x) = 0<br />

Surely you have run into this rule before! How would you derive things like 2u, 3u, 4u, etc.?
 
  • #19
Ah yes that idea just dawned on me in my head, that they go to 0.

So why doesn't u*ln(2) turn into 1*0?
 
  • #20
Because that's not how the product rule works.

Try the real product rule on c*f(x) and you'll see that you get the standard rule of pulling the constant out of the derivative
 
  • #21
Ah I forgot about that... you know for some reason, the last 3 times I refreshed this thread, while the thread was loading the idea which I had forgotten just got posted :P

I think I have a better grasp of this, if I need any extra help I will post.
 

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