<s>[INST] Find the Derivatives of $g(x) = \sqrt{3-2x^2}$ and $h(x) = \ln {(x^2 – x)}$ I am asked to find the derivatives of $g(x) =\sqrt{3-2x^2}$  and $h(x) = \ln{(x^2 – x)} $
For:
$g(x)h(x)$
and
$\dfrac{h(x)}{g(x)}$
and 
$h^3 (x)$
First off I am not sure if my derivatives are correct. Here is what I have..  
$g'(x) = \dfrac{2x} { \sqrt{3 - 2x^2}}$
$h'(x) =\dfrac {2x - 1} {x^2 - x}$
Thanks
Em
 [/INST] Your $h'$ is correct, but your $g'$ should be $\dfrac 12 \cdot \dfrac{-4x}{\sqrt{3-2x^2}}=\dfrac {-2x}{\sqrt{3-2x^2}}$
So from the product rule we have $(h \cdot g)'=h' \cdot g + h \cdot g'=\dfrac {2x-1}{x^2-1} \cdot \sqrt{3-2x^2} + \ln (x^2-x) \cdot \dfrac {-2x}{\sqrt{3-2x^2}}$. You can't really do much to simplify it, except for making the whole second tern negative instead of having $-2x$ in the numerator. 
$\dfrac hg$ is exactly the same, except a different formula.
For $h^3$, we use the chain rule: $(h^3)'=h' \cdot 3(h)^2=\dfrac {2x-1}{x^2-1} \cdot 3(\ln (x^2-x))^2$. Can't do much to simplify this either.
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