Qualification > Math
MAth DOubts
Twinkle Charms:
oh and also i find vectors difficult, if someone can gimme good notes or site from which to refer, Vectors in 2D n 3D? the vector equation n all tht crap :|
Alpha:
--- Quote from: Zara on May 22, 2010, 11:35:20 am ---thankyou alpha, n
its x square multiplied by x-1, so for it we open the brackets n then differientiate? N how to do the diff. part?
--- End quote ---
ln [x^2(x-1)] = ln [x^(2x-2)]
Using laws of indices,
= ln (x^2x/ x^2)
Using laws of logarithms,
= ln (x^2x) - ln (x^2)
Derivative > 2x. x^(2x-1) {which can be simplified to 2x^(2x-1+1)= 2x^2x}/ x^(2x) - 2x/ x^2
= 2x^2x/ x^2x - 2x/ x^2
This one's more complex.
Welcome~
Twinkle Charms:
--- Quote from: ~Alpha on May 22, 2010, 01:40:41 pm ---
ln [x^2(x-1)] = ln [x^(2x-2)]
Using laws of indices,
= ln (x^2x/ x^2)
Using laws of logarithms,
= ln (x^2x) - ln (x^2)
Derivative > 2x. x^(2x-1) {which can be simplified to 2x^(2x-1+1)= 2x^2x}/ x^(2x) - 2x/ x^2
= 2x^2x/ x^2x - 2x/ x^2
This one's more complex.
Welcome~
--- End quote ---
uh-oh seems u neva got wht actually im asking :|
its x(square) this multiplied to (x-1) now get it??
Alpha:
--- Quote from: Zara on May 22, 2010, 01:42:19 pm ---uh-oh seems u neva got wht actually im asking :|
its x(square) this multiplied to (x-1) now get it??
--- End quote ---
Yeah, but isn't that what I did earlier? My first post?
Twinkle Charms:
--- Quote from: ~Alpha on May 22, 2010, 01:45:55 pm ---Yeah, but isn't that what I did earlier? My first post?
--- End quote ---
ooopps my bad, i was reading it rong ur first post...sooorrryyy :-[ and thank youu so mucchh :)
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