<s>[INST] Derivative of $\frac{1}{2s}-\frac{5}{4s^{3}}$ 
$r=\dfrac{1}{2s}-\dfrac{5}{4s^3}$

$r=\dfrac{1}{2}s^{-1}-\dfrac{5}{4}s^{-3}$
$r^{\prime}=-\dfrac{1}{2}s^{-2}-\dfrac{5}{4}(-3)s^{-4}$  
$r^{\prime}=-\dfrac{1}{2s^{2}}+\dfrac{15}{4s^{4}}$
Is this correct? Because I solved it again a different way by using the quotient rule and I got a different answer. Why wouldn't you use the quotient rule here if there are fractions? I found a common denominator and then used the quotient rule and got a different answer. Thanks. 
 [/INST] Finding a common denominator and using the quotient rule we have
$$r=\frac{1}{2s}-\frac{5}{4s^3}$$
$$r=\frac{2s^2-5}{4s^3}$$
$$r'=\frac{4s^3\cdot4s-(2s^2-5)\cdot12s^2}{16s^6}=\frac{16s^4-(24s^4-60s^2)}{16s^6}=\frac{60s^2-8s^4}{16s^6}$$
$$r'=-\frac{1}{2s^2}+\frac{15}{4s^4}$$
As you calculated.
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