In trigonometry, secant function (sec x) is the reciprocal of cosine (cos x) and forms an essential part of calculus. Because tan x = sin x / cos x, having a clear knowledge of these relationships facilitates effective differentiating of sec x. One can use a number of approaches to derive its derivative, some of which include the first principle, quotient rule, and chain rule. In this article, we’ll explore the derivative of sec x, its formula, graphical interpretations, proofs, different differentiation methods, and practical examples to deepen your understanding of this important function.

Derivative of sec x

The ratio of hypotenuse to the adjacent side of an angle in a right triangle is defined as the trigonometric function secant of an angle.

\( \sec \left ( A \right ) = \frac{c}{b} \)

where A denotes the angle, c the hypotenuse, and b the adjacent side

One of the first transcendental functions introduced in Differential Calculus is the Secant Derivative. Secant times tangent, or \sec x .\tan x is the derivative of the secant function (x). where A denotes the angle, c the hypotenuse, and b the adjacent side

This derivative can be proven using limits and trigonometric identities.

\frac{d}{dx}\left ( \sec x \right )

\left ( \sec x \right )’ =\sec x .\tan x

Derivative of Sec X Using First Principle

We will use first principles (or the definition of the derivative to demonstrate that the derivative of sec x is sec x tan x. When attempting to find the derivative, avoid using the first principle unless specifically mentioned in the question. Using the first principle may take longer than the traditional method and may necessitate a good understanding of various forms of formulas related to algebra, trigonometry, and a bit of manipulation. Let’s start.

Assume that \( f\left ( x \right ) = \sec x \)

\( f\left ( x+h \right ) sec\left ( x+h \right ) \)

\( \frac{\mathrm{d} }{\mathrm{d} x} f\left ( x \right )=\lim_{h\rightarrow 0}\left [ \frac{f\left ( x+h \right )-f\left ( x \right )}{h} \right ] \)

\( \frac{\mathrm{d} }{\mathrm{d} x} \left ( \sec x \right ) =\lim_{h\rightarrow 0} \left [ \frac{\sec \left ( x+h \right )-\sec x}{h} \right ] \)

\( \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sec x \right ) =\lim_{h\rightarrow 0} \left [ \frac{1}{h} \left ( \frac{1}{\cos \left ( x+h \right )}-\frac{1}{cos x} \right )\right ] \)

\( \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sec x \right ) =\lim_{h\rightarrow 0} \left [ \frac{1}{h}\left ( \frac{\cos \left ( x \right )- \cos \left ( x+h \right )}{\cos \left ( x+h \right )\cos x} \right ) \right ] \)

\( \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sec x \right ) =\lim_{h\rightarrow 0} \left [ \frac{1}{h} \frac{2 \sin\left ( \frac{2x+h}{2} \sin\left ( \frac{x+h-x}{2} \right )\right )}{h.\cos \left ( x+h \right )\cos x} \right ] \ \)

\( \frac{\mathrm{d}}{\mathrm{d} x}\left ( \sec x \right ) =\lim_{h\rightarrow 0} \left [ \frac{\sin \left ( x+\frac{h}{2} \right )}{\cos \left ( x+h \right )\cos x} \right ] . \lim_{h\rightarrow 0}\left [ \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right ] \)

\( \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sec x \right )=\frac{\sin x}{\cos x.\cos x}\times 1\)

\( \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sec x \right )= \sec x .\tan x \)

Hence, proved.

Derivative of Sec X Using Quotient Rule

Using the quotient rule, we will demonstrate that the differentiation of sec x with respect to x yields sec x tan x.

We will assume that f\left ( x \right )= \sec x, which can be written as f\left ( x \right )= \frac{1}{\cos x}

\( f\left ( x \right ) = \frac{\left [ 1 \right ].\cos -\left [ \cos x \right ].1}{\cos^{2}x} \)

\( = \frac{0.\cos x-\left ( -\sin x \right ).1}{\cos ^{2}x} \)

\( = \frac{0+sin x}{\cos ^{2}x} \)

\( = \frac{sin x}{\cos ^{2}x} \)

\( =\frac{\sin x }{\cos x}.\frac{1}{\cos x} \)

\( =\frac{1}{\cos x}.\frac{\sin x}{\cos x} \)

\(= \sec x.\tan x \)

Hence proved

Derivative of Sec x by Chain Rule

It is , \( \frac{\sin \left ( x \right )}{\cos \left ( x^{2} \right )} \)

\( \sec \left ( x \right ) =\frac{1}{\cos \left ( x \right )} \(\) \frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{\cos \left ( x \right )} =\frac{\mathrm{d} }{\mathrm{d} x} \left ( \cos \left ( x \right )^{-1} \right ) \)

This is equivalent to the chain rule.

\( \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos \left ( x \right )^{-1} \right ) = -\cos \left ( x \right )^{-2} . \)

\( \frac{\mathrm{d} }{\mathrm{d} x} \cos \left ( x \right ) \)

\( = – \frac{1}{\cos \left ( x \right )^{2}} \left ( -\sin \left ( x \right ) \right ) \)

\( = \frac{\sin \left ( x \right )}{\cos \left ( x \right )^{2}} \)

\( = \tan \left ( x \right ) \sec \left ( x \right ) \)

Hence proved

Solved Examples of Derivative of sec x

Problem: 1What is \( \left ( \sec x \right )^{2} \) derivative?

Solution:

\( f\left ( x \right ) = \left ( \sec x \right )^{2} \)

By using power rule and chain rule

\( f’\left ( x \right ) = 2 \sec x\frac{\mathrm{d} }{\mathrm{d} x} \left ( \sec x \right ) \)

\( = 2 \sec x. \left ( \sec x. \tan x \right ) \)

\( = 2 \sec ^{2} x \tan x \)

The derivative of given function is \( = 2 \sec ^{2} x \tan x \)

Problem: 2Determine the derivative of \( \sec ^{-1} x \)

Solution:

\( y = \sec ^{-1} x \)

Then,

\( \sec y = x \) …..(1)

Differentiating both sides in terms of x,

\( \sec y. \tan y\left ( \frac{\mathrm{dy} }{\mathrm{d} x} \right ) =1 \)

\( \frac{\mathrm{dy} }{\mathrm{d} x} =\frac{1}{\sec y.\tan y} \)

According to one of the trigonometric identities,

\( \tan y = \sqrt{\sec ^{2}y-1} = \sqrt{x^{2}-1} \)

\( \frac{\mathrm{dy} }{\mathrm{d} x} = \frac{1}{\left ( \frac{x}{\sqrt{x^{2}-1}} \right )} \)

Derivative of \( \sec ^{-1} x \) is \( \frac{1}{\left ( \frac{x}{\sqrt{x^{2}-1}} \right )} \)

Grasping the derivative of sec x is essential for proficiency in calculus, particularly for the solution of problems relating to rates of change and integrals. Using the rules such as the quotient rule and chain rule, we deduced that d/dx (sec x) = sec x tan x. The result comes into play when performing advanced math involving physics and engineering applications. With a firm understanding of this differentiation process, you can comfortably challenge associated trigonometric derivatives and use them for practical problems, strengthening your total calculus abilities.