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 1

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:

\(\ds \map {\dfrac \d {\d x} } {\sec x}\) \(=\) \(\ds \map {\dfrac \d {\d x} } {\paren {\cos x}^{-1} }\) Exponent Laws
\(\ds \) \(=\) \(\ds \paren {-\sin x} \paren {-\cos^{-2} x}\) Chain Rule for Derivatives, Power Rule
\(\ds \) \(=\) \(\ds \frac 1 {\cos x} \frac {\sin x} {\cos x}\) Exponent Laws
\(\ds \) \(=\) \(\ds \sec x \tan x\) Definition of Real Secant Function and Definition of Real Tangent Function

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

$\blacksquare$


Proof 2

\(\ds \map {\dfrac \d {\d x} } {\sec x}\) \(=\) \(\ds \map {\dfrac \d {\d x} } {\dfrac 1 {\cos x} }\) Definition of Real Secant Function
\(\ds \) \(=\) \(\ds \dfrac {\cos x \map {\frac \d {\d x} } 1 - 1 \map {\frac \d {\d x} } {\cos x} } {\cos^2 x}\) Quotient Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac {0 - \paren {-\sin x} } {\cos^2 x}\) Derivative of Cosine Function, Derivative of Constant
\(\ds \) \(=\) \(\ds \sec x \tan x\) Definition of Real Secant Function, Definition of Real Tangent Function

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

$\blacksquare$


Also see


Sources

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