Answer
$$
y^{\prime}= \frac{d}{dx} \left[ \left(\tan ^{-1} x\right)^{2} \right] =\frac{2 \tan ^{-1} x}{1+x^{2}}
$$
Work Step by Step
$$
y=\left(\tan ^{-1} x\right)^{2}
$$
Differentiating both sides of this equation we have:
$$
\begin{aligned}
\frac{d}{dx} \left[ y \right] &= \frac{d}{dx} \left[ \left(\tan ^{-1} x\right)^{2} \right] \\
y^{\prime}&=2\left(\tan ^{-1} x\right)^{1} \cdot \frac{d}{d x}\left(\tan ^{-1} x\right) \\
&=2 \tan ^{-1} x \cdot \frac{1}{1+x^{2}}\\
&=\frac{2 \tan ^{-1} x}{1+x^{2}}
\end{aligned}
$$
