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**arze** Find in a simplified form, the derivative of the function

$\displaystyle \frac{2+\ln (1+x)^2}{2-\ln (1-x)^2}$

Used the quotient rule:

$\displaystyle \frac{(2-2\ln (1-x))(\frac{2(1)}{(1+x)})-(2+2\ln (1+x))(\frac{-2(-1)}{(1-x)})}{(2-2\ln (1-x))^2}$

$\displaystyle \frac{\frac{4-4\ln(1-x)}{1+x}-\frac{4+4\ln (1+x)}{1-x}}{(2-2\ln(1-x))^2}$

$\displaystyle \frac{\frac{4(1-x)(1-\ln (1-x))-4(1+x)(1+\ln (1+x))}{1-x^2}}{(2-2\ln (1-x))^2}$

$\displaystyle \frac{4(1-x)-4(1-x)(\ln (1-x))-4(1+x)-4(1+x)(\ln (1+x))}{(1-x^2)(2-2\ln (1-x))^2}$

$\displaystyle \frac{4(x-1)(\ln (1-x))-4(1+x)(\ln (1+x))-8x}{(1-x^2)(2-2\ln (1-x))^2}$ (**)

The answer is supposed to be

$\displaystyle \frac{(x-1)\ln (1-x)-(1+x)\ln (1+x)-2x}{(1-x^2)(2-2\ln (1-x))^2}$

so I have a factor 4. Where have i missed out?

Thanks