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B Square root differential problem

  1. Jul 13, 2017 #1
    Hi,

    I working on their text this equation did not make sense to me.

    From equation 1 it differentiate second term , I wonder how he got second term of equation 2.

    What I think is, what I wrote at the bottom

    P_20170713_121813.jpg
     
  2. jcsd
  3. Jul 13, 2017 #2

    Charles Link

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    They are doing a Taylor series and writing ## f(x)=f(a)+f'(a)(x-a) +... ## In this case ## x=\epsilon ## and ## a=0 ##. ## \\ ## I agree with their result.
     
  4. Jul 13, 2017 #3
    but how did they resolve
    f(x)=f(a)+f′(a)(x−a)+..
    to get second term in eq 2.
     
  5. Jul 13, 2017 #4

    Charles Link

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    The chain rule. In numerator of ## f'(\epsilon) ## you have ## 2(y'+\epsilon h) h ##. You then compute ## f'(0) ##.
     
  6. Jul 13, 2017 #5
    but then why there is no ϵh' term in denominator in second term in equation 2 ?
    It should have stayed there.

    and if were to consider ϵ = 0 then why there is still e multiplying second term in equation 2 ?
     
  7. Jul 13, 2017 #6

    Charles Link

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    The ## \epsilon ## is from ## (x-a)=(\epsilon-0)= \epsilon ##. Meanwhile, the ## \epsilon h ## in tthe denominator gets put equal to zero as part of taking ## f'(0) ##. (The entire term is the product of both of these which is ## f'(0)(x-a) ##.)
     
  8. Jul 13, 2017 #7
    So like this ?
     

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    Last edited: Jul 13, 2017
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