Question

24: Show that \(\frac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x\).

Short answer

It is proved that \(\frac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x\).

Step-by-step solution

 Derivative of Trigonometric Functions

The derivative of the trigonometric functions are shown below:

\(\begin{aligned}\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\csc x} \right) &= - \csc x\cot x\\\frac{d}{{dx}}\left( {\cos x} \right) &= - \sin x\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\sec x} \right) &= \sec x\tan x\\\frac{d}{{dx}}\left( {\tan x} \right) &= {\sec ^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\cot x} \right) &= - {\csc ^2}x\end{aligned}\)

Show that \(\frac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x\)

Differentiate the given function with respect to \(x\) as shown below:

\(\begin{aligned}\frac{d}{{dx}}\left( {\sec x} \right) &= \frac{d}{{dx}}\left( {\frac{1}{{\cos x}}} \right)\\ &= \frac{{\cos x\frac{d}{{dx}}\left( 1 \right) - \left( 1 \right)\frac{d}{{dx}}\left( {\cos x} \right)}}{{{{\cos }^2}x}}\\ &= \frac{{\left( {\cos x} \right)\left( 0 \right) - 1\left( { - \sin x} \right)}}{{{{\cos }^2}x}}\\ &= \frac{{\sin x}}{{{{\cos }^2}x}}\\ &= \frac{1}{{\cos x}} \cdot \frac{{\sin x}}{{\cos x}}\\ &= \sec x\tan x\end{aligned}\)

Thus, it is proved that \(\frac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x\).