Derivative of Secant Function

Theorem

 * $\map {\dfrac \d {\d x} } {\sec x} = \sec x \tan x$

where $\cos x \ne 0$.

Proof
From the definition of the secant function:
 * $\sec x = \dfrac 1 {\cos x} = \paren {\cos x}^{-1}$

From Derivative of Cosine Function:
 * $\map {\dfrac \d {\d x} } {\cos x} = -\sin x$

Then:

This is valid only when $\cos x \ne 0$.

Also see

 * Derivative of Sine Function
 * Derivative of Cosine Function


 * Derivative of Tangent Function
 * Derivative of Cotangent Function


 * Derivative of Cosecant Function