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.
Find the Derivative in step by step?
0
$\begingroup$
calculus
derivatives
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0Have you learned the quotient rule for derivatives? The derivative of $\dfrac{f(z)}{g(z)}$ is? – 2012-04-07
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0no! I havn't learn yet. – 2012-04-07
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0Ah. No worries, I think anon's answer should get you there without it. – 2012-04-07
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0Maybe you should first compute the derivative of $z^n$? – 2012-04-07
1 Answers
3
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:
$$\frac{z^2(-4z^{-2}+1-2z)}{z^2}=\frac{-4+z^2-2z^3}{z^2}.$$