anonymous
  • anonymous
derivative of: sqrt(x) - x using the first principal. Please show work. The answer should be 1/[2*sqrt(x)]-1 I cant get the minus 1!
Mathematics
chestercat
  • chestercat
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experimentX
  • experimentX
i guess you know how to find the derivate of x ... so just find for sqrt(x)
anonymous
  • anonymous
f(x) = \[\sqrt{x} -x \]
anonymous
  • anonymous
oh...so just do them one at a time?:| That actually makes a lot of sense. the problem is I have the equation in the "definition of the derivative formula" and its not simplifying down right.

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anonymous
  • anonymous
where you take the limit as h-->0
experimentX
  • experimentX
lim dx->0 (sqrt(x+dx) - sqrt(x))/dx now sqrt(x+dx) = sqrt(x)*sqrt(1+dx/x) using binomial expansion, and removing higher terms --- close to linear expression we have: sqrt(x)(1+1/2*dx/x) or, lim dx->0 (sqrt(x)(1+1/2*dx/x) - sqrt(x))/dx or, lim dx->0 (1/2*dx/sqrt(x))/dx = 1/(2sqrt(x)) hence proved
lgbasallote
  • lgbasallote
seems he wants the increment thingy...the limits
anonymous
  • anonymous
@Ravus , are you asking to take the derivative using the definition? (limit definition of derivative)
perl
  • perl
@experimentX how do you get this sqrt(x+dx) = sqrt(x)*sqrt(1+dx/x)
anonymous
  • anonymous
@perl, nice avatar!
perl
  • perl
hehe
anonymous
  • anonymous
yes I want the "increment thingy"<<
experimentX
  • experimentX
@perl .. take x common!
perl
  • perl
huh?
anonymous
  • anonymous
|dw:1333351654824:dw| plug it in to that...
perl
  • perl
(x+dx)^1/2 = x^1/2 + ...
perl
  • perl
ohhhh
anonymous
  • anonymous
yeah I did plug it in, and I get the right answer almost. I get 1/2*sqrt(x) but that whole thing should have a "-1"! i dont get that bit
perl
  • perl
@experimentX (x+dx)^1/2 = x^1/2 ( 1 + dx/x ) ^1/2
experimentX
  • experimentX
dx is supposed to be del x ... or better h
anonymous
  • anonymous
|dw:1333351804706:dw| do some algebra here....
anonymous
  • anonymous
ok... too many cooks... i'll quit.
anonymous
  • anonymous
yeah I know what the formula is, but the algebra is giving me trouble
perl
  • perl
@experimentX Very clever argument, so you looked at the binomial series and took the first two terms (linear )
experimentX
  • experimentX
yes .. since dx^higer would be very less ... it would seem better to take linear term.
perl
  • perl
then you factored out sqrt x
experimentX
  • experimentX
yes .. and that would cancel out .. sqrt(x) leaving dx/(2sqrt(x)
anonymous
  • anonymous
omg wow finally got it. simple algebra really. lmfao............
experimentX
  • experimentX
congrats
anonymous
  • anonymous
|dw:1333354174909:dw|

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