anyone help me with a basic calculus problem?

Discussion in 'General Discussion' started by Sage, Sep 11, 2006.

  1. Sage

    Sage 13 short of a dozen
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    For some reason I cannot get this darned thing to work out right...

    Differentiate:
    Sq.Rt. (x + sq.rt. (x) )

    It seems to me like it should be:

    1 / [ 4 * sq.rt.(x + sq.rt. (x)) * sq.rt. (x) ]

    BUT, calculator indicates:

    [2 * sq.rt. (x) + 1] / [4 * sq.rt. (x) * sq.rt. {sq.rt. (x) + (sq.rt. (x) + 1) } ]

    I cannot figure out what the heck i'm doing wrong. I'm using chain rule and differentiating whole function, to get 1 / [2 * sq.rt. {x + sq.rt. (x) } ] for the first function, then 1 / [2 * sq.rt. (x) ] for the second one, multiplying them to get the above result. I tried graphing them on the calculator and comparing the slope of the tangent line, as computed by my calculator, of the initial function at x=0.75 and it matches the value of what my calculator indicates if the correct derivative at x=0.75 but it does not match what my work indicates is correct.

    Would greatly appreciate any help.
     
  2. K.I.L.E.R

    K.I.L.E.R Retarded moron
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    To calculate the 7th order PDE...

    I got it:

    g(x) = x + x^0.5
    q(a) = a^0.5

    g'(x) = 1 + 0.5*x^-0.5 = 1 + 1/(2*sqrt(x))
    q'(a) = 0.5*a^-0.5 = 1/2*sqrt(a)

    q'(a)*g'(x) = 1/2*sqrt(a) * 1 + 1/(2*sqrt(x))
    = 1/2*sqrt(x + sqrt(x)) * (1 + 1/2*sqrt(x))

    I love maths.
    Want to play again?
     
  3. pcchen

    pcchen Moderator
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    It should be 1 + 1/(2*sqrt(x)) for the second one.
     
  4. K.I.L.E.R

    K.I.L.E.R Retarded moron
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    I got it first.
    I got it before you did.


     
  5. Sage

    Sage 13 short of a dozen
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    thanks guys, I already figured it out with the help of my mommie :grin:
     
  6. K.I.L.E.R

    K.I.L.E.R Retarded moron
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    Then why ask us if you're going to ask others first? :???:
    I wasted a page in my book for that.
     
  7. Sage

    Sage 13 short of a dozen
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    Well, because I gave upon her helping me but then she called me back.
     
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