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
\[ \displaystyle \dfrac{d}{dx} \sec (u(x)) = \tan u \; \sec u \; \dfrac{d}{dx} u \]
Example 1
Find the derivative of the composite sec functions
\( f(x) = \sec (x^2 + x - 1) \)
\( g(x) = \sec (\sin(x)) \)
\( h(x) = \sec (\sqrt{x+2}) \)
Solution to Example 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
\( \displaystyle \dfrac{d}{dx} f(x) =\tan u \sec u \dfrac{d}{dx} u = \tan (x^2 + x - 1) \sec (x^2 + x - 1) \times (2x + 1) \)
\( = (2x + 1) \; \tan (x^2 + x - 1) \; \sec (x^2 + x - 1) \)
Let \( u(x) = \sin x \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} \sin x = \cos x \) and apply the above rule
\( \displaystyle \dfrac{d}{dx} g(x) = \tan u \sec u \dfrac{d}{dx} u = \tan (\sin x) \sec (\sin x) \times (\cos x) \)
\( = \cos x \; \tan (\sin x) \; \sec (\sin x) \)
Let \( u(x) = \sqrt{x+2} \) and therefore \( \dfrac{d}{dx} u = \dfrac{1}{2\sqrt{x+2}} \) and apply the rule obtained above
\( \displaystyle \dfrac{d}{dx} h(x) = \tan u \sec u \dfrac{d}{dx} u = \tan (\sqrt{x+2}) \sec (\sqrt{x+2}) \times \dfrac{1}{2\sqrt{x+2}} \)
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