Second derivative of $\arctan(x^2)$ Given that $y=\arctan(x^2)$ find $\ \dfrac{d^2y}{dx^2}$.
I got 
$$\frac{dy}{dx}=\frac{2x}{1+x^4}.$$
Using low d high minus high d low over low squared, I got
$$\frac{d^2y}{dx^2}=\frac{(1+x)^4 \cdot 2 - 2x \cdot 4(1+x)^3}{(1+x^4)^2}.$$
I tried to simplify this but didn't get the answer which is
$$\frac{d^2y}{dx^2}=\frac{2(1-3x^4)}{(1+x^4)^2}.$$
Where am I going wrong?

Alternatively, 
$ \large \tan (y) = x^2 \Rightarrow \sec^2 (y) \cdot \frac { \mathrm{d}y}{\mathrm{d}x} = 2x $
$ \large \Rightarrow \frac { \mathrm{d}y}{\mathrm{d}x} = \frac {2x}{\sec^2 (y)} = \frac {2x}{\tan^2 (y) + 1} = \frac {2x}{x^4 + 1} $
$  \large \Rightarrow \frac { \mathrm{d^2}y}{\mathrm{d}x^2} = \frac {2(x^4+ 1) - 2x(4x^3)}{(x^4+1)^2} = \frac {-6x^4 + 2}{(x^4 + 1)^2} $
