Before going to determine the derivative of sec x, let us recall a few things. sec x is the reciprocal of cos x and tan x is the ratio of sin x and cos x. These definitions of sec x and tan x are very necessary to do the differentiation of sec x with respect to x. We can find it using various methods such as:

- by using the first principle
- by using the quotient rule
- by using the chain rule

Let us do the differentiation of sec x in each of these techniques and we will solve a few problems using the derivative of sec x.

Table of Contents

## What is the Derivative of Sec x?

The derivative of sec x with respect to x is sec x · tan x. i.e., it is the product of sec x and tan x. We denote the derivative of sec x with respect to x with d/dx(sec x) (or) (sec x)’. Thus,

**d/dx (sec x) = sec x · tan x (or)****(sec x)’ = sec x · tan x**

But wherever is tan x coming from in the derivative of sec x? We are going to differentiate sec x in various methods such as utilizing the first principles (definition of the derivative), quotient rule, and chain rule in the upcoming sections.

## Derivative of Sec x by First Principle

We are going to show that the derivative of sec x is sec x · tan x by using the first principles (or) the definition of the derivative. For this, assume that f(x) = sec x.

**Proof:**

By first principle, the derivative of a function f(x) is,

f'(x) = limₕ→₀ [f(x + h) – f(x)] / h … (1)

Since f(x) = sec x, we have f(x + h) = sec (x + h).

Substituting these values in (1),

f’ (x) = limₕ→₀ [sec (x + h) – sec x]/h

= limₕ→₀ 1/h [1/(cos (x + h) – 1/cos x)]

= limₕ→₀ 1/h [cos x – cos(x + h)] / [cos x cos(x + h)]

By sum to product formulas, cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2. So

f'(x) = 1/cos x limₕ→₀ 1/h [- 2 sin (x + x + h)/2 sin (x – x – h)/2] / [cos(x + h)]

= 1/cos x limₕ→₀ 1/h [- 2 sin (2x + h)/2 sin (- h)/2] / [cos(x + h)]

Multiply and divide by h/2,

= 1/cos x limₕ→₀ (1/h) (h/2) [- 2 sin (2x + h)/2 sin (- h/2) / (h/2)] / [cos(x + h)]

When h → 0, we have h/2 → 0. So

f'(x) = 1/cos x limₕ/₂→₀ sin (h/2) / (h/2). limₕ→₀ (sin(2x + h)/2)/cos(x + h)

We have limₓ→₀ (sin x) / x = 1. So

f'(x) = 1/cos x. 1. sin x/cos x

We know that 1/cos x = sec x and sin x/cos x = tan x. So

f'(x) = sec x · tan x

Hence proved.

## Derivative of Sec x by Quotient Rule

We will show that the differentiation of sec x with respect to x gives sec x · tan x by using the quotient rule. For this, we will assume that f(x) = sec x and it can be written as f(x) = 1/cos x.

**Proof:**

We have f(x) = 1/cos x = u/v

By quotient rule,

f'(x) = (vu’ – uv’) / v^{2}

f'(x) = [cos x d/dx(1) – 1 d/dx(cos x)] / (cos x)^{2}

= [cos x (0) – 1 (-sin x)] / cos^{2}x

= (sin x) / cos^{2}x

= 1/cos x · (sin x)/(cos x)

= sec x · tan x

Hence proved.

## Derivative of Sec x by Chain Rule

To prove that the derivative of sec x to be sec x · tan x by chain rule, we will assume that f(x) = sec x = 1/cos x.

**Proof:**

We can write f(x) as,

f(x) = 1/cos x = (cos x)^{-1}

By power rule and chain rule,

f'(x) = (-1) (cos x)^{-2} d/dx(cos x)

By a property of exponents, a^{-m} = 1/a^{m}. Also, we know that d/dx(cos x) = – sin x. So

f'(x) = -1/cos^{2}x · (- sin x)

= (sin x) / cos^{2}x

= 1/cos x · (sin x)/(cos x)

= sec x · tan x

Hence proved.

## Solved Examples

**Example 1:** What is the derivative of sec x · tan x?

**Solution:**

Let f(x) = sec x · tan x = uv

By product rule,

f'(x) = uv’ + vu’

= (sec x) d/dx (tan x) + (tan x) d/dx (sec x)

= (sec x)(sec^{2}x) + (tan x) (sec x · tan x)

= sec^{3}x + sec x tan^{2}x

**Answer:** The derivative of the given function is sec^{3}x + sec x tan^{2}x.

**Example 2:** What is the derivative of (sec x)^{2}?

**Solution:**

Let f(x) = (sec x)^{2}.

By power rule and chain rule,

f'(x) = 2 sec x d/dx (sec x)

= 2 sec x · (sec x · tan x)

= 2 sec^{2}x tan x

**Answer:** The derivative of the given function is 2 sec^{2}x tan x.

**Example 3:** Find the derivative of sec^{-1}x.

**Solution:**

Let y = sec^{-1}x.

Then, sec y = x … (1)

Differentiating both sides with respect to x,

sec y · tan y (dy/dx) = 1

dy/dx = 1 / (sec y · tan y)… (2)

By one of the trigonometric identities,

tan y = √sec²y – 1 = √x² – 1

dy/dx = 1/(x √x² – 1)

**Answer:** The derivative of sec⁻¹ x is 1/(x √x² – 1).

## FAQs

**What will be the Derivative of Sec x With Respect to x?**

The derivative of sec x with respect to x is formulated as d/dx(sec x) and it is equal to sec x tan x. i.e., the differentiation of sec x is the product of sec x and tan x.

**How to Find Derivative of Sec x by First Principle?**

The first principle is used to find the derivative of a function f(x) using the formula f'(x) = limₕ→₀ [f(x + h) – f(x)] / h. By substituting f(x) = sec x and f(x + h) = sec (x + h) in this formula and simplifying it, we can see the derivative of sec x to be sec x tan x.

**What is the Differentiation of Sec Square x?**

Sec square x can be written as f(x) = (sec x)^{2}. By power rule and chain rule, its derivative is, f'(x) = 2 sec x d/dx(sec x) = 2 sec x (sec x tan x) = 2 sec^{2}x tan x.

**Is the Derivative of Sec Inverse x the Same as the Derivative of Sec x?**

No, the derivative of sec x is NOT the same as the derivative of sec^{-1}x. The derivative of sec x is sec x tan x whereas the derivative of sec^{-1}x is 1/(x √x² – 1).

**What is the Derivative of Sec x²?**

We know that the derivative of sec w is sec w tan w. Also, by using the chain rule, d/dx (sec x²) = sec x² tan x² d/dx(x²) = 2x sec x² tan x².