# Which derivative rule is this?

1. Aug 15, 2010

### 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 =\)

2. Aug 15, 2010

### rock.freak667

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.

3. Aug 15, 2010

### 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 [Broken]

The only difference between that and the other site is Log vs Ln, which give 2 very different answers.

Last edited by a moderator: May 4, 2017
4. Aug 15, 2010

### Office_Shredder

Staff Emeritus
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

5. Aug 15, 2010

### Xtensity

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?

6. Aug 15, 2010

### Pengwuino

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.

7. Aug 15, 2010

### Xtensity

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.

8. Aug 15, 2010

### Pengwuino

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

9. Aug 15, 2010

### Xtensity

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. Aug 15, 2010

### Office_Shredder

Staff Emeritus
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. Aug 15, 2010

### Xtensity

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

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. Aug 15, 2010

### l'Hôpital

What is d/du u?

13. Aug 15, 2010

### Xtensity

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. Aug 15, 2010

### l'Hôpital

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

15. Aug 15, 2010

### Xtensity

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. Aug 15, 2010

### l'Hôpital

ln 2 is a constant. : )

17. Aug 15, 2010

### Xtensity

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

18. Aug 15, 2010

### 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.?

19. Aug 15, 2010

### 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?

20. Aug 15, 2010

### Office_Shredder

Staff Emeritus
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

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