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I have looked the similar questions but all the questions are hard one and I can't understand.
My question is find the derivative $\dfrac{4z+z^3-z^4} {z^2}$, $z \neq 0$.
I got this answer:
$\dfrac{4+3z^2-4z^3}{z^3}$
however answer is $\dfrac{-2z^3+z^2-4}{z^2}$, looks easy; however, I can't get the answer.
Can you please explain it in step by step? I appreciate it. thanks alot.

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  • $\begingroup$ Have you learned the quotient rule for derivatives? The derivative of $\dfrac{f(z)}{g(z)}$ is? $\endgroup$ – user23784 Apr 7 '12 at 5:52
  • $\begingroup$ no! I havn't learn yet. $\endgroup$ – Cin Sb Sangpi Apr 7 '12 at 5:55
  • $\begingroup$ Ah. No worries, I think anon's answer should get you there without it. $\endgroup$ – user23784 Apr 7 '12 at 5:57
  • $\begingroup$ Maybe you should first compute the derivative of $z^n$? $\endgroup$ – copper.hat Apr 7 '12 at 7:34
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It helps in this case to divide through for a simpler expression:

$$\frac{4z+z^3-z^4}{z^2}=4z^{-1}+z-z^2.$$

Now to differentiate:

$$4(-1)z^{-2}+(1)z^0-(2)z^{1}=-4z^{-2}+1-2z.$$

If you want to put this over a denominator again:

$$\frac{z^2(-4z^{-2}+1-2z)}{z^2}=\frac{-4+z^2-2z^3}{z^2}.$$

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