Using the product rule, we write
[tex]\frac{dy}{dx}=\frac{1}{x^2-2x+2}\times\frac{d(x^2+2x-2)}{dx}+\frac{d(x^2-2x+2)^{-1}}{dx}\times(x^2+2x-2)[/tex][tex]\rightarrow\frac{2x+2}{x^2-2x+2}+\frac{-2(x-1)}{(x^2-2x+2)^2}\times(x^2+2x-2)[/tex]
Next, we find the common denominator of both terms (which is (x^2-2x+2)^2) and then add them.
[tex]\rightarrow\frac{(2x+2)\textcolor{#FF7968}{(x^2-2x+2)}}{(x^2-2x+2)^{\textcolor{#FF7968}{2}}}+\frac{-2(x-1)}{(x^2-2x+2)^2}\times(x^2+2x-2)[/tex][tex]\rightarrow\frac{2x^3-2x^2+4}{(x^2-2x+2)^2}+\frac{-2x^2-2x^2+8x-4}{(x^2-2x+2)^2}[/tex]
And finally, we add the numerators to get
[tex]\rightarrow\frac{-4x^2+8x}{(x^2-2x+2)^2}[/tex]
which is the third choice in the column.
