1. ## More differentiation help

Find in a simplified form, the derivative of the function
$\frac{2+\ln (1+x)^2}{2-\ln (1-x)^2}$

Used the quotient rule:
$\frac{(2-2\ln (1-x))(\frac{2(1)}{(1+x)})-(2+2\ln (1+x))(\frac{-2(-1)}{(1-x)})}{(2-2\ln (1-x))^2}$
$\frac{\frac{4-4\ln(1-x)}{1+x}-\frac{4+4\ln (1+x)}{1-x}}{(2-2\ln(1-x))^2}$
$\frac{\frac{4(1-x)(1-\ln (1-x))-4(1+x)(1+\ln (1+x))}{1-x^2}}{(2-2\ln (1-x))^2}$
$\frac{4(1-x)-4(1-x)(\ln (1-x))-4(1+x)-4(1+x)(\ln (1+x))}{(1-x^2)(2-2\ln (1-x))^2}$
$\frac{4(x-1)(\ln (1-x))-4(1+x)(\ln (1+x))-8x}{(1-x^2)(2-2\ln (1-x))^2}$
The answer is supposed to be
$\frac{(x-1)\ln (1-x)-(1+x)\ln (1+x)-2x}{(1-x^2)(2-2\ln (1-x))^2}$
so I have a factor 4. Where have i missed out?
Thanks

2. Originally Posted by arze
Find in a simplified form, the derivative of the function
$\frac{2+\ln (1+x)^2}{2-\ln (1-x)^2}$

Used the quotient rule:
$\frac{(2-2\ln (1-x))(\frac{2(1)}{(1+x)})-(2+2\ln (1+x))(\frac{-2(-1)}{(1-x)})}{(2-2\ln (1-x))^2}$
$\frac{\frac{4-4\ln(1-x)}{1+x}-\frac{4+4\ln (1+x)}{1-x}}{(2-2\ln(1-x))^2}$
$\frac{\frac{4(1-x)(1-\ln (1-x))-4(1+x)(1+\ln (1+x))}{1-x^2}}{(2-2\ln (1-x))^2}$
$\frac{4(1-x)-4(1-x)(\ln (1-x))-4(1+x)-4(1+x)(\ln (1+x))}{(1-x^2)(2-2\ln (1-x))^2}$
$\frac{4(x-1)(\ln (1-x))-4(1+x)(\ln (1+x))-8x}{(1-x^2)(2-2\ln (1-x))^2}$ (**)
The answer is supposed to be
$\frac{(x-1)\ln (1-x)-(1+x)\ln (1+x)-2x}{(1-x^2)(2-2\ln (1-x))^2}$
so I have a factor 4. Where have i missed out?
Thanks
Who says that's the correct answer? Once you cancel out the 4's from your final answer (** above), then Maple agrees with you!

3. I don't exactly get what you mean. I have the factor 4, but i can't just eliminate it.

4. Originally Posted by arze
$\frac{4(x-1)(\ln (1-x))-4(1+x)(\ln (1+x))-8x}{(1-x^2)(2-2\ln (1-x))^2}$
$(1-x^2)(2-2\ln (1-x))^2 = (1-x^2) 2^2 (1 - \ln(1-x))^2 = 4 (1-x^2) (1 - \ln(1-x))^2$