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

[Xtensity]

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 =\)

 

[rock.freak667]

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

y=2^u⇒lny=ln2^u⇒lny=uln2

Differentiating both sides w.r.t. u

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

In the end, we have

dy/du=2^uln2.

 

[Xtensity]

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.

 

[Office_Shredder]

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

 

[Xtensity]

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?

 

[Pengwuino]

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.

 

[Xtensity]

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.

 

[Pengwuino]

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.

 

[Xtensity]

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?

 

[Office_Shredder]

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

 

[Xtensity]

Ok I understand we use Ln... though I'm having issues with this earlier post:

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

y=2^u⇒lny=ln2^u⇒lny=uln2

Differentiating both sides w.r.t. u

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

In the end, we have

dy/du=2^uln2.

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?

 

[l'Hôpital]

What is d/du u?

 

[Xtensity]

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.

 

[l'Hôpital]

Allow me to rephrase the question. What is the derivative of u with respect to u?

 

[Xtensity]

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?

 

[l'Hôpital]

ln 2 is a constant. : )

 

[Xtensity]

ln 2 is a constant. : )

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

 

[l'Hôpital]

??

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

$$f(x) = c$$

For some constant c, then
$$f'(x) = 0$$

Surely you have run into this rule before! How would you derive things like 2u, 3u, 4u, etc.?

 

[Xtensity]

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?

 

[Office_Shredder]

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

 

[Xtensity]

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.