SteikeTa
Member
+153|6757|Norway/Norwegen/ Norge/Noruega
Have some questions about derivation (correct?).

1.    h(x) = 2xe^x   
Found it to be: h'(x) = 2e^x    Agree?

2     k(x) = 2x * (ln x)^2
      k'(x) = ________  have no idea about this one.

Would be great if someone could show me step by step how to solve this one.
Sydney
2λчиэλ
+783|6853|Reykjavík, Iceland.
Here you go.
https://i43.tinypic.com/2njzhpi.jpg
liquidat0r
wtf.
+2,223|6637|UK
I think you mean differentiation.

1.

Huh? No. You need to use the product rule. You get:

h'(x) = 2*exp(x) + 2*x*exp(x)

http://en.wikipedia.org/wiki/Product_rule

brb, giving a presentation on Atomic Clocks.
Sydney
2λчиэλ
+783|6853|Reykjavík, Iceland.

liquidat0r wrote:

I think you mean differentiation.

1.

Huh? No. You need to use the product rule. You get:

h'(x) = 2*exp(x) + 2*x*exp(x)

http://en.wikipedia.org/wiki/Product_rule

brb, giving a presentation on Atomic Clocks.
Yay! I was right!
Gawwad
My way or Haddaway!
+212|6695|Espoo, Finland
I think the first one is 2xex + 2ex
since I think it's the same as D f(x)g(x) = f(x)g'(x) + g(x)f'(x) ??


Edit:

k(x) = 2x * (ln x)2
follows the same rule

2x * 2(lnx)2-1 * (1/x) + 2 * (ln x)2





(1/x) This is the derivative of ln|x| (i.e. the inner function). Since (ln x)2 is positive it works the same as ln|x|

Last edited by Gawwad (2009-03-12 08:31:20)

SteikeTa
Member
+153|6757|Norway/Norwegen/ Norge/Noruega
Aaaaahhh, damn it!!! The product rule, of course! Jesus, I always fail to remember those rules! So many of them :S
DrunkFace
Germans did 911
+427|6691|Disaster Free Zone

SteikeTa wrote:

Aaaaahhh, damn it!!! The product rule, of course! Jesus, I always fail to remember those rules! So many of them :S
My most useful piece of paper for all of high school maths.
https://img9.imageshack.us/img9/9342/image00010001.jpg

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