Derivative of Arcsecant of Function

Theorem

Let $u$ be a differentiable real function of $x$ such that $\size u > 1$.

Then:

$\map {\dfrac \d {\d x} } {\arcsec u} = \dfrac 1 {\size u \sqrt {u^2 - 1} } \dfrac {\d u} {\d x}$

where $\arcsec$ denotes the arcsecant of $x$.

Proof

\(\ds \map {\frac \d {\d x} } {\arcsec u}\)

\(=\)

\(\ds \map {\frac \d {\d u} } {\arcsec u} \frac {\d u} {\d x}\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \dfrac 1 {\size u \sqrt {u^2 - 1} } \frac {\d u} {\d x}\)

Derivative of Arcsecant Function

$\blacksquare$

Also see

• Derivative of Arcsine of Function
• Derivative of Arccosine of Function

• Derivative of Arctangent of Function
• Derivative of Arccotangent of Function

• Derivative of Arccosecant of Function

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.24$
• 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $17$