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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
\( \displaystyle { \dfrac {d}{dx} \sec x = \dfrac {d}{dx} (\dfrac{ 1 }{\cos x}) = \dfrac { (\dfrac {d}{dx}1) { \cos x } - 1 (\dfrac {d}{dx} \cos x) } {\cos^2 x} } \)

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

\( \displaystyle {= \dfrac{ \sin x } {\cos^2 x} = \dfrac{ \sin x }{\cos x} \dfrac{ 1 }{\cos x} = \tan x \sec x}\)

conclusion
\[ \displaystyle {\dfrac {d}{dx} \sec x = \tan x \; \sec x} \]

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

\( \displaystyle \dfrac{d}{dx} \sec (u(x)) = (\dfrac{d}{du} \sec u) (\dfrac{d}{dx} u ) \)

Simplify

\( = \tan u \sec u \dfrac{d}{dx} u \)

Conclusion

\[ \displaystyle \dfrac{d}{dx} \sec (u(x)) = \tan u \; \sec u \; \dfrac{d}{dx} u \]

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.