<s>[INST] CAL I - Quotient Rule to find Derivative I'm stuck on this problem $h(x) = \frac{\sqrt{x}}{x^3+1}$.
I get to here: $\frac{x^3+1-x^{1/2}(3x^2)}{2x^{1/2}(x^3+1)^2}$.
I started going this route:  $\frac{x^3+1-3x^{5/2}}{2x^{1/2}(x^3+1)^2}$.
The book's solution is getting this in their step process: $\frac{x^3+1-6x^3}{2x^{1/2}(x^3+1)^2}$.
How are they getting the $-6x^3$ in the numerator? 
 [/INST] Using 
$$
\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}
$$ we have 
$$
\left(\frac{\sqrt{x}}{x^3+1}\right)'=\frac{\frac{1}{2\sqrt{x}}(x^3+1)-\sqrt{x}(3x^2)}{(x^3+1)^2}=\frac{1}{2\sqrt{x}}\frac{(x^3+1)-2x\:(3x^2)}{(x^3+1)^2}=\frac{1}{2\sqrt{x}}\frac{-5x^3+1}{(x^3+1)^2}
$$ as desired.
Hoping it's clear for you now.
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