<s>[INST] Find the derivative of $\frac{5-\frac{1}{x^2}}{x+3}$ Test 2 Review #4
Find the  derivative of $\frac{5-\frac{1}{x^2}}{x+3}$
$Solution$:
The first step to calculating this derivative is to rewrite it as so:
$\frac{5-x^{-2}}{x+3}$
Now we will apply the quotient rule:
$\frac{d}{dx}(\frac{5-x^{-2}}{x+3})$
$=(\frac{(x+3)(\frac{d}{dx}(5-x^{-2}))-(5-x^{-2})(\frac{d}{dx}(x+3))}{(x+3)^2})$
$=(\frac{(x+3)((\frac{d}{dx}5-\frac{d}{dx}x^{-2}))-(5-x^{-2})((\frac{d}{dx}x+\frac{d}{dx}3))}{(x+3)^2})$
$=(\frac{(x+3)((0-(-2x^{-3}))-(5-x^{-2})((1)+0))}{(x+3)^2})$
$=(\frac{(x+3)(2x^{-3})-(5-x^{-2})}{(x+3)^2})$
$=(\frac{(2x^{-2}+6x^{-3})-5+x^{-2})}{(x+3)^2})$
$=(\frac{2x^{-2}+6x^{-3}-5+x^{-2})}{(x+3)^2})$
$=(\frac{3x^{-2}+6x^{-3}-5}{(x+3)^2})$
This is an acceptable answer. We could multiply the top and bottom by $x^3$:
$=(\frac{3x^{-2}+6x^{-3}-5}{(x+3)^2})(\frac{x^3}{x^3})$
$=(\frac{3x+6-5x^3}{x^3(x+3)^2})$
 [/INST] I like the following way.
$$\left(\frac{5-\frac{1}{x^2}}{x+3}\right)'=\frac{\frac{2}{x^3}}{x+3}+\left(5-\frac{1}{x^2}\right)\left(-\frac{1}{(x+3)^2}\right)=$$
$$=\frac{2(x+3)-5x^3+x}{x^3(x+3)^2}=\frac{3x+6-5x^3}{x^3(x+3)^2},$$ which gives the same result. 
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