Find the second derivative using the quotient rule

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Homework Statement


\frac{d^2}{dz^2} [ (z^2) \(\frac{-(z-\frac{1}{z})^2}{20+8(z+\frac{1}{z})} \frac{1}{iz}]


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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?
 
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It is often easier to simplify first, and then differentiate. Let's see if we can rewrite your product of fractions:

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

which finally simplifies to

i \cdot \frac{z^4 + 2z + 1}{8z^2+20z+8} =<br /> \frac{i}{4} \cdot \frac{z^4+2z^2+1}{2z^2+5z+2}

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).<br /> \frac{i}{4} \cdot \frac{z^4+2z^2+1}{2z^2+5z+2} =<br /> \frac{i}{4} (z^4+2z+1) \cdot (2z^2+5z+2)^{-1}<br />

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).
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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