Derivative of secx – Derivation, Explanation, and Example

Our discussion will focus on the derivative of $\boldsymbol{\sec x}$. We’ll also make sure that we understand how we can derive this rule. After the discussion, we should be able to differentiate other functions containing secant.

The derivative of $\boldsymbol{\sec x}$ returns the product of $\boldsymbol{\sec x}$ and $\boldsymbol{\tan x}$. We can prove this derivative by differentiating $\boldsymbol{\dfrac{1}{\cos x}}$.

Here’s a breakdown of the concepts that we’ll learn in this article:

  • Apply the rule for $\dfrac{d}{dx} \sec x$.
  • Learn how to incorporate this to differentiate trigonometric functions that contain it.
  • Know how we can use fundamental and trigonometric derivative rules to differentiate secant.

Are you ready? Let’s begin by learning how to use the derivative of secant and what this means for us.

What is the derivative of secx?                

The derivative of $\sec x$ is simply the product of $\boldsymbol{\sec x}$ and $\boldsymbol{\tan x}$ as shown below.

\begin{aligned}\dfrac{d}{dx} \sec x = \sec x \tan x\end{aligned}

We can use this rule to differentiate functions such as the ones shown below.

\begin{aligned}f(x) &= \sec (2x -1)\\g(x) &= \cos (2x) – \sec \left(\dfrac{1}{x – 1}\right)\\ y&= \sec(\sqrt{2x – 1}) \end{aligned}

Make sure to keep the chain rule in mind when differentiating functions like these. This way, we can account for the inner functions.

How to find the derivative of secx and functions containing it?

Recall that $\sec x$ is simply the reciprocal of $\cos x$, so we can instead differentiate $\dfrac{1}{\cos x}$ to derive the formula for $\dfrac{d}{dx} \sec x$. We can then apply the quotient rule to differentiate $\dfrac{1}{\cos x}$ as shown below.

\begin{aligned}\dfrac{d}{dx} \sec x &=\dfrac{d}{dx} \dfrac{1}{\cos x}\\&= \dfrac{\cos x \dfrac{d}{dx}(1) – 1\dfrac{d}{dx} (\cos x)}{(\cos x)^2}, \phantom{x}\color{DarkOrange}\text{Quotient Rule}\\&= \dfrac{\cos x {\color{DarkOrange}(0)} – 1{\color{Purple}(-\sin x)}}{(\cos x)^2}, \phantom{x}{\color{DarkOrange}\text{Constant Rule}}\phantom{x}\text{&}{\color{Purple}\text{ Derivative of Cosine}}\end{aligned}

Simply this expression and use trigonometric identities to rewrite the resulting derivative.

\begin{aligned}\dfrac{d}{dx} \sec x &= \dfrac{0 + \sin x}{(\cos x)^2}\\&= \dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x}\\&= \sec x \tan x\end{aligned}

This shows that $\dfrac{d}{dx} \sec x$ is indeed equivalent to $\sec x \tan x$.

Now that we understand how $\dfrac{d}{dx} \sec x$ was derived, let’s see how we can apply this derivative rule. For example, we can use the derivative of $\sec x$ to differentiate $h(x) = \sec (3x + 4)$. Of course, we’ll have to use the chain rule first to differentiate $h(x)$.

\begin{aligned}h'(x) &= \dfrac{d}{dx} f[g(x)] \\&= f'[g(x)] \cdot g'(x)\\f(x) &= \sec x\\ g(x) &= 3x +4\\h'(x) &= \sec (3x + 4)\end{aligned}

We can use the derivative of $\sec x$, $\dfrac{d}{dx} \sec x = \sec x \tan x$ to find $f’(x)$. We can apply the fundamental derivative rules to differentiate the inner function, $g(x)$.

\begin{aligned}h'(x)&= {\color{DarkOrange}\sec[g(x)]\tan [g(x)]} \cdot g'(x),\phantom{x} {\color{DarkOrange} \text{Derivative of Secant}}\\&= \sec(3x +4)\tan(3x +4) \cdot g'(x)\\&=\sec(3x +4)\tan(3x +4)\cdot\color{DarkOrange}\left(\dfrac{d}{dx}3x+\dfrac{d}{dx}4\right),\phantom{x} {\color{DarkOrange} \text{Sum Rule}} \\&=\sec(3x +4)\tan(3x +4)\cdot\left({\color{DarkOrange}3\dfrac{d}{dx}x}+{\color{Purple}0}\right),\phantom{x} {\color{DarkOrange} \text{Constant Multiple Rule }}\text{&}{\color{Purple} \text{ Constant Rule}}\\&=\sec(3x +4)\tan(3x +4)\cdot\left(3({\color{DarkOrange}1})+ 0\right),\phantom{x} {\color{DarkOrange} \text{Power Rule }}\\&=3\sec(3x +4)\tan(3x +4)\end{aligned}

This shows that $\dfrac{d}{dx} \sec(3x +4) = 3\sec(3x +4)\tan(3x + 4)$. We’ve prepared more examples for you to work on to master this derivative rule. Check them out once you’re ready!

Example 1

Find the derivative of $h(x) = \sec \sqrt{x + 2}$.


The function $h(x)$ is a composite function that has $f(x) = \sec x$ and $g(x) = \sqrt{x  + 2}$ as the outer and inner functions, respectively. This means that we’ll have to apply the chain rule first.

\begin{aligned}h'(x) &= \dfrac{d}{dx} f[g(x)] \\&= f'[g(x)] \cdot g'(x)\\f(x) &= \sec x\\ g(x) &= \sqrt{x + 2}\\h'(x) &= \sec (\sqrt{x + 2})\end{aligned}

We can use the derivative rule, $\dfrac{d}{dx} \sec x = \sec x \tan x$, to differentiate the outer function. The second function, $g(x) = \sqrt{x + 2}$, will also require the chain rule but we’ll breeze through this to focus on our main function.

\begin{aligned}h'(x)&= {\color{DarkOrange}\sec[g(x)]\tan [g(x)]} \cdot g'(x),\phantom{x} {\color{DarkOrange} \text{Derivative of Secant}}\\&= \sec(\sqrt{x + 2})\tan(\sqrt{x + 2}) \cdot g'(x)\\&=\sec(\sqrt{x + 2})\tan(\sqrt{x + 2})\cdot\left[{\color{DarkOrange}\dfrac{1}{2}(x + 2)^{\frac{1}{2}-1}}\cdot {\color{Purple}\dfrac{d}{dx}(x + 2)} \right],\phantom{x} {\color{DarkOrange} \text{Power Rule }}\text{&}{\color{Purple} \text{ Chain Rule}}\\&= \sec(\sqrt{x + 2})\tan(\sqrt{x + 2})\cdot\left[\dfrac{1}{2\sqrt{x + 2}}\cdot \left(\dfrac{d}{dx}x + \dfrac{d}{dx}2 \right )\right],\phantom{x} {\color{DarkOrange} \text{Sum Rule }}\end{aligned}

Simplify this expression further to find the expression for $h’(x)$.

\begin{aligned}h'(x)&= \sec(\sqrt{x + 2})\tan(\sqrt{x + 2})\cdot\left[\dfrac{1}{2\sqrt{x + 2}}\cdot \left({\color{DarkOrange}1} + {\color{Purple}0}\right )\right],\phantom{x} {\color{DarkOrange} \text{Power Rule }}\text{&}\color{Purple}\text{ Constant Rule}\\&= \dfrac{1}{2\sqrt{x + 2}}\sec(\sqrt{x + 2})\tan(\sqrt{x + 2})\\&= \dfrac{\sec(\sqrt{x + 2})\tan(\sqrt{x + 2})}{2\sqrt{x + 2}}\end{aligned}

Hence, we have $h’(x) = \dfrac{\sec(\sqrt{x + 2})\tan(\sqrt{x + 2})}{2\sqrt{x + 2}}$.

Example 2

Find the derivative of $h(x) = \dfrac{\sec x}{\sqrt{x – 4}}$.


This time, $h(x)$ is a rational expression that contains $\sec x $ in its numerator and $\sqrt{x – 4}$ in its denominator. We can use the quotient rule, $\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x)\dfrac{d}{dx}f(x) – f(x)\dfrac{d}{dx}g(x)}{[g(x)]^2}$, to start differentiating $h(x)$.

\begin{aligned}h'(x)&= \dfrac{\sqrt{x – 4}\dfrac{d}{dx}(\sec x) – \sec x\dfrac{d}{dx}\sqrt{x – 4}}{(\sqrt{x – 4})^2}\\&=\dfrac{\sqrt{x – 4}\dfrac{d}{dx}(\sec x) – \sec x\dfrac{d}{dx}\sqrt{x – 4}}{x – 4}\end{aligned}

We can use the derivative rule for $\sec x $ to simplify the first term in the numerator. We can apply the power and chain rules to differentiate $\sqrt{x + 4}$.

\begin{aligned}\boldsymbol{\dfrac{d}{dx}\sec x}\end{aligned}

\begin{aligned}\boldsymbol{\dfrac{d}{dx}\sqrt{x + 4}}\end{aligned}

\begin{aligned}\dfrac{d}{dx} \sec x &= \sec x \tan x \end{aligned}

\begin{aligned}\dfrac{d}{dx} \sqrt{x + 4}&= {\color{DarkOrange}\dfrac{1}{2}(\sqrt{x +4})^{\frac{1}{2} – 1}} \cdot {\color{Purple}\dfrac{d}{dx}(x + 4)}, \phantom{x}{\color{DarkOrange}\text{Power Rule }}\text{&}{\color{Purple}\text{ Chain Rule}}\\&= \dfrac{1}{2\sqrt{x +4}}\left({\color{DarkOrange}\dfrac{d}{dx}x + \dfrac{d}{dx}4 }\right ),\phantom{x}{\color{DarkOrange}\text{Sum Rule}}\\&=\dfrac{1}{2\sqrt{x + 4}}({\color{DarkOrange}1} + {\color{Purple}0}),\phantom{x}{\color{DarkOrange}\text{Power Rule }}\text{&}{\color{Purple}\text{ Constant Rule}}\\&= \dfrac{1}{2\sqrt{x +4}} \end{aligned}

Let’s use these expressions to simplify the numerator of $h’(x)$ as shown below.

\begin{aligned}h'(x)&=\dfrac{(\sqrt{x – 4})(\sec x\tan x)- (\sec x)\left(\dfrac{1}{2\sqrt{x + 4}} \right )}{x – 4}\\&= \dfrac{\dfrac{1}{2\sqrt{x – 4}}[2(x – 4)(\sec x \tan x)] – \sec x}{x – 4}\\&= \dfrac{(\sec x)[2(x – 4)\tan x – 1]}{\sqrt{(x – 4)^3}}\end{aligned}

This means that $h’(x) = \dfrac{(\sec x)[2(x – 4)\tan x – 1]}{\sqrt{(x – 4)^3}}$.

Practice Questions

1. Find the derivative of the following functions.
a. $f(x) = \sec(2x + 5)$
b. $g(x) = \sec(x^2 – 2x+1)$
c. $h(x) = \sec\left(\dfrac{1}{x}\right)$
2. Find the derivative of the following functions.
a. $f(x) = \sec(\sqrt{4x – 1})$
b. $g(x) = \sec(\sqrt{x^2 + 4x +4})$
c. $h(x) = \sec\left(\dfrac{x + 1}{x – 3}\right)$
3. Find the derivative of the following functions.
a. $f(x) = \dfrac{e^{4x – 1}}{\sec x}$
b. $g(x) = \dfrac{\sec(3x)}{\ln x}$
c. $h(x) = \sec\left(\sqrt{\dfrac{x}{x^2 – 1}}\right)$

Answer Key

a.$f’(x) = 2\sec(2x +5)\tan(2x + 5)$
b.$g’(x) = 2(x -1) \sec(x^2 – 2x+1) \tan(x^2 – 2x+1)$
c. $h’(x) =-\dfrac{1}{x^2}\sec\left(\dfrac{1}{x}\right) \tan\left(\dfrac{1}{x}\right)$
a. $f’(x) = \dfrac{2\sec(\sqrt{4x – 1})\tan(\sqrt{4x – 1})}{\sqrt{4x – 1}}$
b. $g’(x) = \dfrac{(x + 2)\sec(\sqrt{x^2 + 4x+4})\tan(\sqrt{ x^2 + 4x+4})}{\sqrt{ x^2 + 4x + 4}}$
c. $h’(x)= -\dfrac{4\sec\left(\dfrac{x + 1}{x – 3}\right) \tan\left(\dfrac{x + 1}{x – 3}\right)}{(x -3)^2}$
a. $f’(x) = e^{4x – 1}(\sec x)(\tan x +4)$
b. $g’(x) = \dfrac{[\sec (3x)][3x (\ln x)\tan(3x) – 1]}{x (\ln x)^2}$
c. $h’(x) = -\dfrac{(x^2 + 1) \sec\left(\sqrt{\dfrac{x}{x^2 – 1}\tan\left(\sqrt{\dfrac{x}{x^2 – 1}}\right)}\right)}{2\sqrt{\dfrac{x}{x^2 – 1}}(x^2 – 1)^2}$

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