Before going to find 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 important to do the differentiation of sec x with respect to x. We can find it using various ways such as:

by using the first principleby using the quotient ruleby using the chain rule

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

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1. | What is Derivative of Sec x? |

2. | Derivative of Sec x by First Principle |

3. | Derivative of Sec x by Quotient Rule |

4. | Derivative of Sec x by Chain Rule |

5. | FAQs on Derivative of Sec x |

## What is 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 where is tan x coming from in the derivative of sec x? We are going to differentiate sec x in various methods such as using 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 prove 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.

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**Proof:**

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

f'(x) = limₕ→₀

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

Substituting these values in (1),

f' (x) = limₕ→₀

= limₕ→₀ 1/h <1/(cos (x + h) - 1/cos x)>

= limₕ→₀ 1/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> /

= 1/cos x limₕ→₀ 1/h <- 2 sin (2x + h)/2 sin (- h)/2> /

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)> /

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 prove 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') / v2

f'(x) =

=

= (sin x) / cos2x

= 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/am. Also, we know that d/dx(cos x) = – sin x. So

f'(x) = -1/cos2x · (- sin x)

= (sin x) / cos2x

= 1/cos x · (sin x)/(cos x)

= sec x · tan x

Hence proved.

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