The derivative of sec x with respect to x is equal to secx tanx. The secx is the reciprocal of cosx. In this post, we will learn how to find the derivatives of sec x using the following methods:

- First principle of derivatives
- Chain rule of derivatives
- Quotient rule of derivatives.

## What is the Derivative of Sec x?

The derivative of secx with respect to x is denoted by the symbol $\frac{d}{dx}$(sec x) or (sec x)$’$ and it is equal to secx tanx. Using the fact $\sec x =\frac{1}{\cos x}$, we can find the derivative of sec x by the chain rule and quotient rule of derivatives.

## Derivative of Sec x Formula

The formula for the derivative of secx is given below.

- $\frac{d}{dx}$(sec x) = sec x tan x, (or)
- (sec x)$’$ = sec x tan x.

## Derivative of Sec x by First Principle

We will show that the differentiation of sec x is sec x tan x by the first principle. Putting f(x) = sec x in the derivative formula by first principle $f'(x)$ $=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$, we get the derivative of sec x. Therefore,

## Derivative of Sec x by Chain Rule

By the chain rule, we will now show that the differentiation of sec x is sec x tan x. We have

$\sec x =\dfrac{1}{\cos x}$

Put t=cosx. So $\dfrac{dt}{dx}=-\sin x$. Now,

$\dfrac{d}{dx}(\sec x)=\dfrac{d}{dx}(\dfrac{1}{\cos x})$

= $\dfrac{d}{dt}(\dfrac{1}{t}) \cdot \dfrac{dt}{dx}$, by the chain rule of derivatives

= $\dfrac{d}{dt}$(t^{-1}) ⋅ (-sinx)

= -t^{-1-1} ⋅ (-sinx) by the power rule of derivatives

= $\dfrac{\sin x}{\cos^2 x}$ as t=cosx.

= secx tanx.

So the derivative of secx by chain rule is secx tanx.

**Also Read:**

Derivative of tanx | The derivative of tanx is sec^{2}x. |

Derivative of cotx | The derivative of cotx is -cosec^{2}x. |

Derivative of sin^{2}x | The derivative of sin^{2}x is 2sinx cosx. |

Derivative of e^{cosx} | The derivative of e^{cosx} is -sinx e^{cosx}. |

## Derivative of Sec x by Quotient Rule

We will now find the derivative of secx by the quotient rule. This rule is used to find the derivative of a quotient function which says that

$(\dfrac{f}{g})’=\dfrac{gf’-fg’}{g^2}$, where $’$ denotes the first order derivative.

Let us express secx as a quotient function as follows.

$\sec x =\dfrac{1}{\cos x}$

Applying the above quotient rule of derivatives with f=1 and g=cosx, we get that

$(\sec x)’ =\dfrac{\cos x \cdot 1^\prime – 1 \cdot (\cos x)’}{\cos^2 x}$

= $\dfrac{\cos x \cdot 0 – 1 \cdot (-\sin x)}{\cos^2 x}$

= $\dfrac{\sin x}{\cos^2 x}$ = secx tanx.

So the derivative of secx by quotient rule is secx tanx.

## FAQs on Derivative of Sec x

**Q1: what is the derivative of secx?**

Answer: The derivative of secx is secx tanx. This can proved using the first principle, chain rule, and quotient rule of derivatives.

**Q2: Is secx differentiable at the point x=0?**

Answer: As the derivative of secx is secx tanx = sinx/cos^{2}x and cosx is undefined at x=0, the function secx is not differentiable at x=0.

**Q3: What is the integration of secx?**

Answer: The integration of secx is ln|secx+tanx|+C where C is an integration constant and ln is the logarithm with base e, called the natural logarithm.