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I am having trouble figuring out if I can simplify this expression anymore. This is the most I was able to simplify it with. Let me know if anyone has ideas please! Thanks

$$ \frac{(x^4 \cos x)+(4x^3 \sin x)}{(x^4 \sin x)} $$

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    $\begingroup$ It's helpful to use Mathjax when you are writing your questions, as it is a little unclear what your question means. When you write (x^4cosx), do you mean (a) $x^4 \times \cos x$ or (b) $x^{4\cos x}$? $\endgroup$
    – Joe
    May 29 '20 at 20:35
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$$ ((x^4 \cos x)+(4x^3 \sin x))/(x^4 \sin x)\\ (x^4 \cos x)/(x^4 \sin x)+(4x^3 \sin x)/(x^4 \sin x)\\ \cot x + 4/x $$

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Just do it:

$\frac {x^4 \cos x + 4x^3 \sin x}{x^4 \sin x} = \frac {x^4\cos x}{x^4\sin x} + \frac {4x^3 \sin x}{x^4 \sin x} = \frac {\cos x}{\sin x} + \frac 4x$.

If you like we can replace $\frac {\cos x }{\sin x} = \cot x$ to get $\cot x + \frac 4x$ but we do not have to.

And we can replace $\frac 1x$ with $x^{-1}$ if we want to get $\cot x + 4x^{-1}$ but again we don't have to.

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Hint: let $a=x^4 \cos x, b=4x^3 \sin x, c=x^4 \sin x$ ,For $a, b \in \mathbb{R}$ and $c\in \mathbb{R^{*}}$ we have: $\displaystyle\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

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