Answer
$f''(x) = \frac{-(4 x ((1 + x^2) - x sin(2 x)) sec^2(x))}{(1 + x^2)^3} + \frac{(sec^2(x) (2 x - sin(2 x) - 2 x cos(2 x)))}{(1 + x^2)^2}+ \frac{(2 ((1 + x^2) - x sin(2 x)) tan(x) sec^2(x))}{(1 + x^2)^2}$
Work Step by Step
a) Using WolframAlpha and inputting "limit (tan(x+h)/(1+(x+h)^2) - tan(x)/(1+x^2))/h as h approaches 0", we get $f'(x) =\frac{sec^2(x) (1 + x^2 - x sin(2 x))}{(1 + x^2)^2}$
b) $f'(x) = \frac{(x^2+1)sec^2(x) - tan(x)(2x)}{(x^2+1)^2} = \frac{(sec^2(x) (1 + x^2 - x sin(2 x)))}{(1 + x^2)^2}$
c) Using WolframAlpha and inputting "derivative (sec^2(x) (1 + x^2 - x sin(2 x)))/(1 + x^2)^2", we get $f''(x) = \frac{-(4 x ((1 + x^2) - x sin(2 x)) sec^2(x))}{(1 + x^2)^3} + \frac{(sec^2(x) (2 x - sin(2 x) - 2 x cos(2 x)))}{(1 + x^2)^2}+ \frac{(2 ((1 + x^2) - x sin(2 x)) tan(x) sec^2(x))}{(1 + x^2)^2}$
