Second derivative of arctan(x)

Recall that $f'(x) = \dfrac{1}{1+x^2} = (1+x^2)^{-1}$. This is the first derivative that we now need to differentiate again to get $f''(x)$.

Apply the chain rule to $(1+x^2)^{-1}$: bring down the exponent and reduce it by one, then multiply by the derivative of the inside function $1+x^2$, which is $2x$. This gives $f''(x) = -1\cdot(1+x^2)^{-2}\cdot 2x = \dfrac{-2x}{(1+x^2)^2}$.

The second derivative is negative x over the quantity one plus x squared, all squared. Since the denominator is always positive, the sign depends only on negative x. For a positive value of x, negative x is negative, so the second derivative is negative.