Find y''' for y = arctan x/a, a ≠0
Find $y'''$ for $y = \arctan\dfrac{x}{a}$, $a\neq0$.
Explanation of Method
After the first step the arctangent is gone and only a rational function remains.
Step 1: First derivative
\[ y' = \frac{1}{1+\dfrac{x^{2}}{a^{2}}}\cdot\frac1a = \frac{a^{2}}{a^{2}+x^{2}}\cdot\frac1a = \frac{a}{a^{2}+x^{2}}. \]
From here on the problem is purely rational.
Step 2: Second derivative
\[ y' = a\bigl(a^{2}+x^{2}\bigr)^{-1} \;\Longrightarrow\; y'' = a\cdot\bigl(-1\bigr)\bigl(a^{2}+x^{2}\bigr)^{-2}\cdot2x = \frac{-2ax}{\bigl(a^{2}+x^{2}\bigr)^{2}}. \]
Step 3: Third derivative
Quotient rule, with $\bigl[\bigl(a^{2}+x^{2}\bigr)^{2}\bigr]' = 4x\bigl(a^{2}+x^{2}\bigr)$: \[y''' = \frac{-2a\bigl(a^{2}+x^{2}\bigr)^{2}+2ax\cdot4x\bigl(a^{2}+x^{2}\bigr)}{\bigl(a^{2}+x^{2}\bigr)^{4}}.\]
One factor $\bigl(a^{2}+x^{2}\bigr)$ cancels: \[y''' = \frac{-2a\bigl(a^{2}+x^{2}\bigr)+8ax^{2}}{\bigl(a^{2}+x^{2}\bigr)^{3}} = \frac{-2a^{3}-2ax^{2}+8ax^{2}}{\bigl(a^{2}+x^{2}\bigr)^{3}} = \frac{6ax^{2}-2a^{3}}{\bigl(a^{2}+x^{2}\bigr)^{3}}.\]
Step 4: Verification
\[x = 0:\ y''' = \frac{2a\bigl(-a^{2}\bigr)}{a^{6}} = -\frac{2}{a^{3}} \;\;\checkmark\]
\[a = 1,\ x = 1:\ y''' = \frac{2\cdot2}{8} = \frac12; \quad\text{numerically } \frac{y''(1.001)-y''(0.999)}{0.002} \approx 0.5000 \;\;\checkmark\]
\[y''' = 0 \text{ at } x = \pm\frac{a}{\sqrt3}, \text{ the inflection points of } y'' \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y''' = \dfrac{2a\bigl(3x^{2}-a^{2}\bigr)}{\bigl(a^{2}+x^{2}\bigr)^{3}}\,}\]
Remark
Remark. The function $y' = \dfrac{a}{a^{2}+x^{2}}$ is, up to a constant, the Cauchy density, and its repeated derivatives are what produce the Chebyshev-like numerators seen here. Note that $y''$ vanishes only at $x = 0$, where $y$ itself has its inflection: the arctangent bends one way for $x\lt 0$ and the other for $x>0$.