Question 200907
I'm assuming that the problem is *[Tex \LARGE y=\frac{\ln(x+2)}{x^2+1}] right???



Recall that if *[Tex \LARGE y=\frac{f(x)}{g(x)}], then 


*[Tex \LARGE y^{\prime}=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{(g(x))^2}]



In your case, *[Tex \LARGE y=\frac{\ln(x+2)}{x^2+1}], which means that *[Tex \LARGE f(x)=\ln(x+2)] and *[Tex \LARGE g(x)=x^2+1]



Now derive f and g to get: *[Tex \LARGE f^{\prime}(x)=\frac{1}{x+2}] and *[Tex \LARGE g^{\prime}(x)=2x]




Now plug these functions into the formula *[Tex \LARGE y^{\prime}=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{(g(x))^2}] to get:



*[Tex \LARGE y^{\prime}=\frac{\left(\frac{1}{x+2}\right)(x^2+1)-\ln(x+2)(2x)}{(x^2+1)^2}]



Now multiply every term by the inner LCD x+2 and simplify



*[Tex \LARGE y^{\prime}=\frac{x^2+1-2x(x+2)\ln(x+2)}{(x+2)(x^2+1)^2}]




you could perform more simplifications, but that would only change how the answer looks.