Prove the inverse-tangent derivative by implicit differentiation

Let $y=\tan^{-1}x$, so $\tan y=x$ with $-\pi/2<y<\pi/2$. Differentiating gives $\sec^2y\,y^{\prime}=1$, hence $y^{\prime}=1/\sec^2y$.

$\sec^2y=1+\tan^2y=1+x^2$, so $y^{\prime}=1/(1+x^2)$, as printed. The denominator is positive for every real $x$, and the principal-range tangent is defined throughout the corresponding open interval.