Derivative of Secant of Function
Theorem
Let $u$ be a differentiable real function of $x$.
Then:
- $\map {\dfrac \d {\d x} } {\sec u} = \sec u \tan u \dfrac {\d u} {\d x}$
where $\sec$ is the secant function and $\tan$ is the tangent function.
Proof
| \(\ds \map {\frac \d {\d x} } {\sec u}\) | \(=\) | \(\ds \map {\frac \d {\d u} } {\sec u} \frac {\d u} {\d x}\) | Chain Rule for Derivatives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sec u \tan u \frac {\d u} {\d x}\) | Derivative of Secant 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: $13.18$
- 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $7$