Derivative of Secant Function

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Theorem

$D_x \left({\sec x}\right) = \sec x \tan x$

where $\cos x \ne 0$.


Proof

From the definition of the secant function:

$\sec x = \dfrac 1 {\cos x} = \left({\cos x}\right)^{-1}$

From Derivative of Cosine Function:

$D_x \left({\cos x}\right) = -\sin x$


Then:

\(\displaystyle D_x \left({\sec x}\right)\) \(=\) \(\displaystyle D_x \left({ \left({\cos x}\right)^{-1} }\right)\) $\quad$ Exponent Laws $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({-\sin x}\right) \left({-\cos^{-2} x}\right)\) $\quad$ Chain Rule, Power Rule $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\cos x} \frac {\sin x} {\cos x}\) $\quad$ Exponent Laws $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sec x \tan x\) $\quad$ Definitions of secant and tangent $\quad$

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

$\blacksquare$


Sources