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$$F(p) = \frac{\cos p + pe^p}{p^{10} + \dfrac{1}{p^2}}$$

O.K. What a simple function! /s

By quotient law we have:

$$\frac{dF}{dp}=\frac{(-\sin(p) + pe^p + e^p)(p^{10} + p^{-2}) - (\cos(p) + pe^p)(10p^9 - 2p^{-3})}{(p^{10} + p^{-2})^2}$$

Some fractional simplification later:

$$\frac{(\sin(-p) + pe^p + e^p)(\frac{p^{12} + 1}{p^2}) - (\cos(p) + pe^p)(\frac{10p^{12} - 2}{p^3})}{p^{20} + p^{-4} + 2p^{12}}$$

Still very long, still not very jolly. No matter how much I think I can't think of a way to simplify it further. Can anyone prompt me? (Emphasis on prompt, please don't give me the entire solution).

Thanks in advance!

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  • $\begingroup$ For instance, multiply numerator and denominator by $p^4$, so that you cancel the fractions in the numerator and you remove negative exponents in the denominator. $\endgroup$ – AugSB Mar 11 '16 at 17:37
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Rewrite it as $$ F(p)=\frac{p^2\cos p+p^3e^p}{p^{12}+1}=\frac{f(p)}{g(p)} $$ and apply the quotient rule: \begin{align} f'(p)&=2p\cos p-p^2\sin p+3p^2e^p+p^3e^p\\[4px] g'(p)&=12p^{11} \end{align} Therefore \begin{align} f'(p)g(p) &=(2p\cos p-p^2\sin p+3p^2e^p+p^3e^p)(p^{12}+1)\\[4px] &=2p(p^{12}+1)\cos p-p^2(p^{12}+1)\sin p+(3p^2+p^3)(p^{12}+1)e^p \end{align} and $$ f(p)g'(p)= 12p^{13}\cos p+12p^{14}e^p $$ Now, just put together the pieces

$f'(p)g(p)-f(p)g'(p)=2p(1-5p^{12})\cos p-p^2(p^{12}+1)\sin p+p^2(3+p-9p^{11}+p^{12})e^p$

and, finally,

$F'(p)=\dfrac{2p(1-5p^{12})\cos p-p^2(p^{12}+1)\sin p+p^2(3+p-9p^{11}+p^{12})e^p}{(p^{12}+1)^2}$

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  • $\begingroup$ When I used @AugSB's comment, I got: $$\dfrac{(p^{14} + p^2)\sin(-p) + (10p^{13} - 2p)\cos(p) + (11p^{14} + p^{15} + p^3 - p^2)e^p}{p^{24} + p + 2p^{16}}$$ $\endgroup$ – naiveai Mar 12 '16 at 10:20
  • $\begingroup$ @eshansingh1 Why multiplying by $p^4$? Just $p^2$ suffices (which is essentially the same I did). $\endgroup$ – egreg Mar 12 '16 at 10:21
  • $\begingroup$ Yes, but what I'm more worried about is the disagreeing coefficient for $e^p$. I'm not sure how to check this. Forgive me, I'm not really a math person. $\endgroup$ – naiveai Mar 12 '16 at 10:24
  • $\begingroup$ @eshansingh1 Such computations are always a mess and it's quite easy making some small mistake. I used all the care I could in my answer, but some mistake could have sneaked in nonetheless. $\endgroup$ – egreg Mar 12 '16 at 10:25
  • $\begingroup$ Yeah, you're right. Let me just have a look through it again. $\endgroup$ – naiveai Mar 12 '16 at 10:27
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Rewrite your function as $$ F(p) = \mathrm{Re}\left[p^2(e^{\mathrm{i}p} + pe^p)(p^{12}+1)^{-1}\right]\ , $$ where Re stands for real part. Differentiating (which is fast using product and chain rule), you will get a sum of three reasonable (at least to my taste) fractions.

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  • $\begingroup$ Sorry if this is a bit of a stupid question, but... where did the cosine go in your rewrite? And where did $i$ come from? $\endgroup$ – naiveai Mar 12 '16 at 3:15
  • $\begingroup$ @eshansingh1 Ok, I assume you haven't seen Euler's formula before en.wikipedia.org/wiki/Euler's_formula ... in this case, ignore my answer. $\endgroup$ – Pierpaolo Vivo Mar 12 '16 at 9:43
  • $\begingroup$ Yep, though it is very interesting. Thanks for that! $\endgroup$ – naiveai Mar 12 '16 at 10:06
  • $\begingroup$ You're welcome. $\endgroup$ – Pierpaolo Vivo Mar 12 '16 at 12:16

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