1. quick Q about differentiating

Hi, this is kinda urgent so please respond quickly if you can.

My "root" function (dunno if it's called that in English, I'm from Austria) is:

f(x) = 2x • lnx

My first "derivative" (?) is:

f'(x) = 2 • lnx + 2

Now my question is what the second derivative would be. I don't know how to apply the rules here since I have a multiplication and an addittion there.

Is this correct:

f''(x) = 2/x + lnx

Thanks...

2. Originally Posted by temhawk
My first "derivative" (?) is:

f'(x) = 2 • lnx + 2

Now my question is what the second derivative would be. I don't know how to apply the rules here since I have a multiplication and an addittion there.

Is this correct:

f''(x) = 2/x + lnx
No, the second derivative is 2/x. (Both of the 2s in 2 • lnx + 2 are constants, so their derivative is 0.)

3. Hi, thanks for your reply. I don't understand why it is 2/x. The rules for differentiating are:

(Products)
f(x) = u • v ........ f'(x) = u' • v + u • v'

(Sums)
f(x) = u + v ........ f'(x) = u' + v'

My comprehension problem is: how are these two rules combined when you have something like:

f(x) = u • v + s

(Like I have^^)

I would think:

f'(x) = u' • v + u • v' + s'

That would give me:

f'(x) = lnx + 2 • (1/x) + 0 = 2/x + lnx

What am I missing??? (I have the feeling it's something to do with constants vs variables but I can't figure it out.)

Thanks...

Edit: Actually, I think I know how you did it. In my formula reference there is a rule like this:

y = k • f(x) ........ y' = k • f'(x) CONSTANT FACTOR IS KEPT

That works when you consider the first "2" (from f(x) = 2 • lnx + 2) to be "k" and the rest ("lnx + 2") f(x). But there is another formula right above that one in my reference book that goes like this:

y = f(x) + k ........ y' = f'(x) CONSTANT SUMMAND IS OMITTED

And "2 • lnx" could be considered as "f(x)" and the last "2" as the summand ("k"), then you would get what I got. So is this a matter of which rule takes precedence? If so, when do I use the rule, with which I would get my (wrong) result in my current function here?

Thanks...

4. Originally Posted by temhawk
Hi, thanks for your reply. I don't understand why it is 2/x. The rules for differentiating are:

(Products)
f(x) = u • v ........ f'(x) = u' • v + u • v'

(Sums)
f(x) = u + v ........ f'(x) = u' + v'

My comprehension problem is: how are these two rules combined when you have something like:

f(x) = u • v + s

(Like I have^^)
Okay, you're trying to differentiate 2 • ln x + 2. So u = 2, v = ln x and s = 2.

Originally Posted by temhawk
I would think:

f'(x) = u' • v + u • v' + s'

That would give me:

f'(x) = lnx + 2 • (1/x) + 0 = 2/x + lnx
If u = 2 then u' = 0. So f'(x) = 0 • ln x + 2 • (1/x) + 0 = 2/x.

5. Ah, it's clear to me now! So I can use both formulae I found in my reference interchangeably as long as I don't forget that differentiated constants are zero and not just nothing =P

And my own combo-formula works too!

Thanks!!