Derivative of Secant Function

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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:

\(\displaystyle \map {\dfrac \d {\d x} } {\sec x}\) \(=\) \(\displaystyle \map {\dfrac \d {\d x} } {\paren {\cos x}^{-1} }\) Exponent Laws
\(\displaystyle \) \(=\) \(\displaystyle \paren {-\sin x} \paren {-\cos^{-2} x}\) Chain Rule for Derivatives, Power Rule
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\cos x} \frac {\sin x} {\cos x}\) Exponent Laws
\(\displaystyle \) \(=\) \(\displaystyle \sec x \tan x\) Definitions of secant and tangent

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

$\blacksquare$


Also see