<s>[INST] minima of $\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}$ Can anyone help me in finding minima of $\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}$ where $k$ and $k'$, are constants. I found the differential but it was too big to be equated.
$\frac{dy}{dx}=\frac{(1-k)\log(x^2-x)[\ln x^2(x-k')x+2(1-k')(x-1)]-(1-k,)\ln x^2[\ln(x^2-x)(x-1)(1-k+x)+(2x-1)(1-k)}{((1-k')(x-1)\log x^2)^2}$
 [/INST] Wolfram gave an easier derivative.  
$$
\dfrac{d}{dx}\left(\frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right) = \dfrac{((k-1) ((2 x-1) \log(x^2)-\log((x-1) x) (\log(x^2)+2 x-2)))}{((k'-1) (x-1)^2 \log^2(x^2))}
$$
To see this result visit this link. 
Setting the numerator of the derivative equal to $0$,
$$
(k-1) ((2 x-1)\log(x^2)-\log((x-1) x) (\log(x^2)+2 x-2))=0
$$
We get the roots:
$$
x_1 \approx -2.5915\\
x_2 \approx  -0.352004\\
x_3 \approx  2.94663
$$
Evaluating $x_1$: 
$$
\left. \frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right| _{x_1} \approx\frac{(0.845206 k-0.845206)}{(k'-1)}
$$
Evaluating $x_2$:
$$
\left. \frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right| _{x_2} \approx \frac{(0.092577 k-0.092577)}{(k'-1)}
$$
Evaluating $x_3$:
$$
\left. \frac{(1-k)x\log(x^2-x)}{(1-k')(x-1)\log x^2}\right| _{x_3} \approx\frac{(1.22336 k-1.22336)}{(k'-1)}
$$
So $x_2$ appears to be satisfactory. 
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