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Homework Help: Find the derivative by hand.

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex] \frac{d^2}{dz^2} [ (z^2) \(\frac{-(z-\frac{1}{z})^2}{20+8(z+\frac{1}{z})} \frac{1}{iz}] [/tex]


    2. Relevant equations



    3. The attempt at a solution

    I have to do this by hand no calculators, CAS, or tables. I started by just expanding things and using the quotient rule, but it got real messy real fast. Any ideas?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 3, 2007 #2
    It is often easier to simplify first, and then differentiate. Let's see if we can rewrite your product of fractions:

    [tex]z^2\cdot \frac{-(z-\frac{1}{z})^2}{20+8(z+\frac{1}{z})} \cdot \frac{1}{iz}
    = \frac{z^2[-(z^2-2+\frac{1}{z^2})]}{20+8(z+\frac{1}{z})} \cdot \frac{1}{iz}
    = \frac{-(z^4 + 2z^2 + 1)}{[20+8(z+\frac{1}{z})]z} \cdot \frac{1}{i}
    = -\frac{z^4 + 2z^2 + 1}{20z+8z^2+8} \cdot \frac{1}{i}
    = -\frac{z^4+2z+1}{8z^2+20z+8} \cdot (-i)
    [/tex]

    which finally simplifies to

    [tex]i \cdot \frac{z^4 + 2z + 1}{8z^2+20z+8} =
    \frac{i}{4} \cdot \frac{z^4+2z^2+1}{2z^2+5z+2} [/tex]

    You can stop here and use the quotient rule for the first derivative; then differentiate the result to obtain the second derivative.

    You may find it easier to write this fraction as a product (I personally consider the product rule to be cleaner and faster than the quotient rule, though they are equivalent in many respects).


    [tex]
    \frac{i}{4} \cdot \frac{z^4+2z^2+1}{2z^2+5z+2} =
    \frac{i}{4} (z^4+2z+1) \cdot (2z^2+5z+2)^{-1}
    [/tex]

    Using this form, just differentiate using the product rule, and don't simplify the result (because it makes finding the second derivative easier; we just apply the product rule again).
     
    Last edited: Dec 4, 2007
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