Derivative of Arctangent of Function

Theorem

Let $u$ be a differentiable real function of $x$.

Then:

$\map {\dfrac \d {\d x} } {\arctan u} = \dfrac 1 {1 + u^2} \dfrac {\d u} {\d x}$

where $\arctan$ denotes the arctangent of $x$.


Proof

\(\ds \map {\frac \d {\d x} } {\arctan u}\) \(=\) \(\ds \map {\frac \d {\d u} } {\arctan u} \frac {\d u} {\d x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac 1 {1 + u^2} \frac {\d u} {\d x}\) Derivative of Arctangent Function

$\blacksquare$


Also see


Sources

  • 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.22$
  • 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $15$
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