Derivative of sec x

The derivative of \( \sec (x)\) is calculated using the quotient rule of derivatives.

Proof of the Derivative of sec x

A trigonometric identity relating \( \sec x \) and \( \cos x \) is given by \[ \sec x = \dfrac { 1 }{ \cos x } \] We use the quotient rule of differentiation to find the derivative of \( \sec x \); hence \[ \dfrac {d}{dx} \sec x = \dfrac {d}{dx} \left(\dfrac{ 1 }{\cos x} \right) = \dfrac { \left(\dfrac {d}{dx}1 \right) { \cos x } - 1 \left(\dfrac {d}{dx} \cos x\right) } {\cos^2 x} \] The derivative of the constant 1 is equal to zero. Use the formulae for the derivative of the trigonometric functions \( \cos x \) given by \( \dfrac {d}{dx}\cos x = - \sin x \) and substitute to obtain \[ \displaystyle {\dfrac {d}{dx} \sec x = \dfrac{ (0 - (-\sin x) )}{\cos^2 x}} \] Simplify \[ {= \dfrac{ \sin x } {\cos^2 x} = \dfrac{ \sin x }{\cos x} \dfrac{ 1 }{\cos x} = \tan x \sec x}\] conclusion
\[ \boxed{ {\dfrac {d}{dx} \sec x = \tan x \; \sec x} } \]

Graph of sec x and its Derivative

The graphs of \( \sec(x) \) and its derivative are shown below.

Graph of sec x and its derivative

Derivative of the Composite Function sec (u(x))

We now consider the composite function sec of another function u(x). Use the chain rule of differentiation to write \[ \dfrac{d}{dx} \sec (u(x)) = (\dfrac{d}{du} \sec u) (\dfrac{d}{dx} u ) \] \[ = \tan u \sec u \dfrac{d}{dx} u \] Conclusion \[ \boxed{ \dfrac{d}{dx} \sec (u(x)) = \tan u \; \sec u \; \dfrac{d}{dx} u} \]

Example 1
Find the derivative of the composite sec functions

  1. \( f(x) = \sec (x^2 + x - 1) \)
  2. \( g(x) = \sec (\sin(x)) \)
  3. \( h(x) = \sec (\sqrt{x+2}) \)

Solution to Example 1


  1. Let \( u(x) = x^2 + x - 1 \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} (x^2 + x - 1) = 2x + 1 \) and apply the rule for the composite sec function given above

    \( \displaystyle \dfrac{d}{dx} f(x) =\tan u \sec u \dfrac{d}{dx} u = \tan (x^2 + x - 1) \sec (x^2 + x - 1) \times (2x + 1) \)

    \( = (2x + 1) \; \tan (x^2 + x - 1) \; \sec (x^2 + x - 1) \)


  2. Let \( u(x) = \sin x \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} \sin x = \cos x \) and apply the above rule

    \( \displaystyle \dfrac{d}{dx} g(x) = \tan u \sec u \dfrac{d}{dx} u = \tan (\sin x) \sec (\sin x) \times (\cos x) \)

    \( = \cos x \; \tan (\sin x) \; \sec (\sin x) \)


  3. Let \( u(x) = \sqrt{x+2} \) and therefore \( \dfrac{d}{dx} u = \dfrac{1}{2\sqrt{x+2}} \) and apply the rule obtained above

    \( \displaystyle \dfrac{d}{dx} h(x) = \tan u \sec u \dfrac{d}{dx} u = \tan (\sqrt{x+2}) \sec (\sqrt{x+2}) \times \dfrac{1}{2\sqrt{x+2}} \)

    \( = \dfrac {\tan (\sqrt{x+2}) \; \sec (\sqrt{x+2})} {{2\sqrt{x+2}} } \)


More References and links

Rules of Differentiation of Functions in Calculus.
Trigonometric Identities and Formulas.
Derivatives of the Trigonometric Functions.
Chain Rule of Differentiation in Calculus.