Derivative of Arccotangent of Function
Jump to navigation
Jump to search
Theorem
Let $u$ be a differentiable real function of $x$.
Then:
- $\map {\dfrac \d {\d x} } {\arccot u} = -\dfrac 1 {1 + u^2} \dfrac {\d u} {\d x}$
where $\arccot$ denotes the arccotangent of $x$.
Proof
\(\ds \map {\frac \d {\d x} } {\arccot u}\) | \(=\) | \(\ds \map {\frac \d {\d u} } {\arccot u} \frac {\d u} {\d x}\) | Chain Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 {1 + u^2} \frac {\d u} {\d x}\) | Derivative of Arccotangent 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: $13.23$
- 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $16$