Derivative of Secant Function/Proof 1

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:

\(\ds \map {\dfrac \d {\d x} } {\sec x}\)

\(=\)

\(\ds \map {\dfrac \d {\d x} } {\paren {\cos x}^{-1} }\)

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 $\cos x \ne 0$.

$\blacksquare$