Derivative of Secant Function/Proof 2
Jump to navigation
Jump to search
Theorem
- $\map {\dfrac \d {\d x} } {\sec x} = \sec x \tan x$
where $\cos x \ne 0$.
Proof
\(\ds \map {\dfrac \d {\d x} } {\sec x}\) | \(=\) | \(\ds \map {\dfrac \d {\d x} } {\dfrac 1 {\cos x} }\) | 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 $\cos x \ne 0$.
$\blacksquare$
Proof
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Quotient