Derivative of f(x)=ln((x^2)+1): I got (2x)/(x^2 + 1) and was wondering....

[Shade080]

So I had to find the derivative of f(x)=ln((x^2)+1) and got (2x)/((x^2)+1) and was wondering why you can't simplify it to 2/(x+1)? I checked and the answer is (2x)/((x^2)+1).

 

[stapel]

So I had to find the derivative of f(x)=ln((x^2)+1) and got (2x)/((x^2)+1) and was wondering why you can't simplify it to 2/(x+1)?

You have the derivative as being:

. . . . .\(\displaystyle f'(x)\, =\, \dfrac{2x}{x^2\, +\, 1}\)

How are you proposing to factor an "x" out of "x² + 1", in order to have a common factor to cancel? (Scroll down almost halfway here, and start reading at "Warning:".)