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y'=(x^3 - 2x)^ln(x)lnx^3-2x+lnx*(2x-2)/(x^3-2x)

I forgot an 1/x next to (x^3 - 2x)

multiplication

I got\[\Large y'=[\frac{(3x^2-2)\ln(x)}{x^3-2x}+\frac{\ln(x^3-2x)}{x}](x^3-2x)^{\ln(x)}\]

hm

yes

thank you guys