Give single-fraction derivatives of a log radical and inverse tangent

Rewrite as $\frac12\ln(a^2-x^2)$. Its derivative is $\frac12\dfrac{-2x}{a^2-x^2}=\dfrac{x}{x^2-a^2}$ on the real domain $a^2-x^2>0$.

The source labels this route not recommended: $\dfrac1{\sqrt{a^2-x^2}}\cdot\frac12(a^2-x^2)^{-1/2}(-2x)=-\dfrac{x}{a^2-x^2}=\dfrac{x}{x^2-a^2}$.

The chain rule gives $\dfrac1{1+(1/(2x))^2}\left(-\dfrac2{(2x)^2}\right)=-\dfrac2{4x^2+1}$ for $x\ne0$.