Having said this, probably the sine and the cosine functions come to your mind when talking about trigonometric functions, and maybe the tangent function as well. But we have a total of six trigonometric functions! It is time to give some screen time to the secant, the cosecant, and the cotangent functions.

The secant function, along with the cosecant and the cotangent functions, are collectively known as the **reciprocal functions** because they are the reciprocal of the main trigonometric functions. Here you will learn how to find the derivative of each.

## Derivative of the Secant Function sec

The secant function is the reciprocal of the cosine function.

The **secant function** is denoted as

\[\sec{x}\]

and is the reciprocal of the **cosine function**, that is

\[\sec{x}=\frac{1}{\cos{x}}.\]

To find the derivative of the secant function you can use the derivative of the cosine function and the quotient rule. Begin by writing the secant function in terms of the cosine function, that is

\[\sec{x}=\frac{1}{\cos{x}}.\]

Next, you can use the quotient rule, so

\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \sec{x} &= \frac{ \left(\frac{\mathrm{d}}{\mathrm{d}x}(1)\right)(\cos{x}) - (1)\left(\frac{\mathrm{d}}{\mathrm{d}x}\cos{x}\right) }{(\cos{x})^2} \\ &= \frac{ \left(\frac{\mathrm{d}}{\mathrm{d}x}(1)\right)(\cos{x}) - (1)\left(\frac{\mathrm{d}}{\mathrm{d}x}\cos{x}\right) }{\cos^2{x}}. \end{align}\]

The derivative of the constant \(1\) is equal to \(0\), and the derivative of the cosine function is the negative of the sine function,

\[ \frac{\mathrm{d}}{\mathrm{d}x} \cos{x} = -\sin{x},\]

so

\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\sec{x} &= \frac{(0)(\cos{x}) -(1)(-\sin{x} ) }{\cos^2{x}} \\ &= \frac{\sin{x}}{\cos^2{x}} \\ &= \left( \frac{1}{\cos{x}} \right) \left(\frac{\sin{x}}{\cos{x}} \right).\end{align}\]

For the last step, you can rewrite the reciprocal of the cosine as the secant again, and you can also use the trigonometric identity

\[\frac{\sin{x}}{\cos{x}}=\tan{x},\]

obtaining

\[ \frac{\mathrm{d}}{\mathrm{d}x}\sec{x} = (\sec{x})(\tan{x}).\]

You will usually find the above expression in tables of derivatives, just written without parentheses. This gives you the formula for the derivative of the secant function.

The derivative of the secant function is

\[ \frac{\mathrm{d}}{\mathrm{d}x}\sec{x} = \sec{x}\,\tan{x}.\]

## Derivative of the Cotangent Function cot

It is time to move on to the cotangent function, which is the reciprocal of the tangent function.

The **cotangent function** is denoted as

\[\cot{x}\]

and is the reciprocal of the **tangent function,** that is

\[\cot{x}=\frac{1}{\tan{x}}.\]

One particularity of the tangent and the cotangent functions is that they can also be written as rational functions using the sine and the cosine functions, as seen in one of the steps required to find the derivative of the secant function. For the tangent function, you can write

\[\tan{x}=\frac{\sin{x}}{\cos{x}}.\]

Since the cotangent function is the reciprocal of the tangent function, you can also find the cotangent function written as a rational function using the sine and the cosine functions as well, that is

\[\begin{align} \cot{x} &= \frac{1}{\tan{x}} \\ &= \frac{1}{\frac{\sin{x}}{\cos{x}}}. \end{align}\]

Using properties of fractions, you can write this as

\[\cot{x}=\frac{\cos{x}}{\sin{x}},\]

which means that the cotangent function can also be written as the quotient of the cosine function and the sine function.

You can use the above identity to find the derivative of the cotangent function. Since it is a quotient of two functions, you will need to use the Quotient Rule, so

\[ \frac{\mathrm{d}}{\mathrm{d}x} \cot{x} = \frac{\left(\frac{\mathrm{d}}{\mathrm{d}x}\cos{x}\right) (\sin{x}) - (\cos{x}) \left( \frac{\mathrm{d}}{\mathrm{d}x} \sin{x} \right) }{(\sin{x})^2}.\]

Next, you will need to use the fact that the derivative of the sine function is the cosine function,

\[ \frac{\mathrm{d}}{\mathrm{d}x}\sin{x} = \cos{x}, \]

and you previously found the derivative of the cosine function. By substituting these derivatives, you will get

\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\cot{x} &= \frac{(-\sin{x})(\sin{x})-(\cos{x})(\cos{x})}{(\sin{x})^2} \\ &= -\frac{\sin^2{x}+\cos^2{x}}{\sin^2{x}}. \end{align}\]

Here you will also need to use the Pythagorean identity,

\[\sin^2{x}+\cos^2{x}=1,\]

so

\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \cot{x} &= -\frac{\sin^2{x}+\cos^2{x}}{\sin^2{x}} \\&= - \frac{1}{\sin^2{x}}. \end{align} \]

Finally, use the properties of the exponents to write

\[ \frac{\mathrm{d}}{\mathrm{d}x} \cot{x} = - \left(\frac{1}{\sin{x}}\right)^2. \]

As you will see further below, the reciprocal of the sine function is the cosecant function, \( \csc{x}\), so you can write

\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \cot{x} &= -(\csc{x})^2 \\ &=-\csc^2{x}. \end{align}\]

This gives you the formula for the derivative of the cotangent function.

The derivative of the cotangent function is

\[ \frac{\mathrm{d}}{\mathrm{d}x} \cot{x} = -\csc^2{x}.\]

## Derivative of the Cosecant Function csc

Finally, you will also find the reciprocal of the sine function.

The **cosecant function** is denoted as

\[\csc{x}\]

and is the reciprocal of the **sine function**, that is

\[\csc{x}=\frac{1}{\sin{x}}.\]

You can find the derivative of the cosecant function just like you did with the secant function. Begin by writing the cosecant function in terms of the sine function,

\[ \csc{x} = \frac{1}{\sin{x}}.\]

Next, differentiate using the Quotient Rule, that is

\[ \frac{\mathrm{d}}{\mathrm{d}x} \csc{x} = \frac{ \left( \frac{\mathrm{d}}{\mathrm{d}x} (1) \right) (\sin{x}) - (1) \left( \frac{\mathrm{d}}{\mathrm{d}x}\sin{x} \right) }{(\sin{x})^2}, \]

where you can use the derivative of the sine function, so

\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \csc{x} &= \frac{(0)(\sin{x})-(1)(\cos{x})}{\sin^2{x}} \\ &= -\frac{\cos{x}}{\sin^2{x}} \\ &= -\left(\frac{1}{\sin{x}} \right) \left( \frac{\cos{x}}{\sin{x}} \right).\end{align}\]

Finally, rewrite the reciprocal and use the cotangent function, so

\[\frac{\mathrm{d}}{\mathrm{d}x}\csc{x} = -(\csc{x})(\cot{x}).\]

Once again, you will most likely find the formula written without parentheses.

The derivative of the cosecant function is

\[\frac{\mathrm{d}}{\mathrm{d}x}\csc{x}=-\csc{x}\,\cot{x}.\]

## Derivative of the Inverse Secant Function

You have seen that the secant function is the **reciprocal** of the cosine function. However, you might be wondering how to deal with the **inverse** secant function.

The inverse secant function, also known as the arcus secant function, is denoted as

\[\mathrm{arcsec}{\,x}\]

and is the **inverse function** of the secant function.

You might also find the inverse secant function written as

\[\sec^{-1}{x},\]

where you have to keep in mind that the \(-1\) is **not** an exponent, it is used to denote an inverse function.

Do not forget that inverse is **not **the same as reciprocal**.**

Whenever you are talking about inverse functions, you need to be careful with their domain. For the inverse secant function you have to consider that the outputs of the secant function are such that\[ |\sec{x}| \geq 1, \]

so the domain of the inverse secant function will be all numbers whose absolute value is greater than or equal to \(1\), that is

\[ (-\infty,-1] \cup [1,\infty).\]

Also, since the secant function is a periodic function, it is possible to obtain the same output from two different inputs. In order to make sure that the inverse secant is a function, this range has to be restricted and the usual convention is that its outputs are between \(0\) and \(\pi\), except \(\frac{\pi}{2}\), so

\[ 0 \leq \mathrm{arcsec}{\, x} \leq \pi, \text{where,}\, \mathrm{arcsec}{\, x} \neq \frac{\pi}{2}.\]

After sorting this out, it is time to look at the derivative of the inverse secant function, which can be obtained using implicit differentiation and some trigonometric identities.

The derivative of the inverse secant function is

\[\frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arcsec}{\,x} = \frac{1}{|x|\sqrt{x^2-1}}.\]

## Derivative of the Inverse Cotangent Function

The inverse cotangent function, as its name suggests, is the inverse of the cotangent function.

The inverse cotangent function, also known as the arcus cotangent function, is denoted as

\[\mathrm{arccot}{\, x}\]

and is the inverse function of the cotangent function.

An alternative notation for the inverse cotangent function is

\[\cot^{-1}{x}.\]

You can obtain any real number as an output of the cotangent function, so the domain of the inverse cotangent function consists of all real numbers.

The outputs of the inverse cotangent function are usually chosen such that they lie between \(0\) and \(\pi\), not including these values. This means that

\[ 0 < \mathrm{arccot}{\, x} < \pi.\]

Please note that some books might define the range between \( -^\pi/_2 \) and \( ^\pi/_2,\) without including \(0\), that is

\[ \left[-\frac{\pi}{2},0 \right) \cup \left( 0, \frac{\pi}{2} \right].\]

The derivative of the inverse cotangent function is

\[ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccot}{\, x} = -\frac{1}{x^2+1}\]

## Derivative of the Inverse Cosecant Function

Do not forget about the inverse cosecant function!

The inverse cosecant function, also known as the arcus cosecant function, is denoted as

\[\mathrm{arccsc}{\, x}\]

and is the inverse function of the cosecant function.

The inverse cosecant function can also be written as

\[ \csc^{-1}{x},\]

and its derivative differs from the derivative of the inverse secant by a sign.

Just like the inverse secant function, the outputs of the cosecant function are such that\[ |\csc{x}| \geq 1, \]

so the domain of the inverse cosecant function will be all numbers whose absolute value is greater than or equal to \(1\), that is

\[ (-\infty,-1] \cup [1,\infty).\]

The outputs of the inverse cosecant function are such that they are between \( -^\pi/_2\) and \( ^\pi/_2\), except \(0\). That is

\[ -\frac{\pi}{2} \leq \mathrm{arccsc}{\,x} \leq \frac{\pi}{2}, \text{where,}\, \mathrm{arccsc}{\, x} \neq 0.\]

The derivative of the inverse cosecant function is

\[\frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccsc}{\, x} = -\frac{1}{|x|\sqrt{x^2-1}}.\]

To see how this and the rest of the derivatives are obtained, please take a look at our article on the Derivatives of Inverse Trigonometric Functions.

## Examples of Derivatives of sec, csc, and cot

Practice the above derivatives by doing some examples!

Find the derivative of

\[ f(x) = \sec{2x^2}.\]

**Solution:**

To find this derivative you will need to use the Chain Rule along with the Power Rule and the derivative of the secant function. Begin by letting \[ u=2x^2,\]

so the Chain Rule tells you that

\[ f'(x)= \frac{\mathrm{d}}{\mathrm{d}u}\sec{u} \frac{\mathrm{d}u}{\mathrm{d}x}.\]

By using the Power Rule, you obtain

\[ \frac{\mathrm{d}u}{\mathrm{d}x} = 4x,\]

so

\[ f'(x)= \left( \frac{\mathrm{d}}{\mathrm{d}u} \sec{u} \right) (4x), \]

now use the derivative of the secant function, giving you

\[ f'(x) = (\sec{u}\,\tan{u}) (4x).\]

Finally, substitute back \(u\) and rearrange, that is

\[ f'(x) = (4x) \sec{2x^2} \, \tan{2x^2}.\]

You can also use the Product Rule to find derivatives of the reciprocal trigonometric functions!

Find the derivative of

\[ g(x) = x\cot{x}.\]

**Solution:**

Here you will need to use the Product Rule, that is

\[ g'(x) = \left(\frac{\mathrm{d}}{\mathrm{d}x} x \right) \cot{x} + x\left( \frac{\mathrm{d}}{\mathrm{d}x}\cot{x} \right).\]

Next, use the Power Rule and the derivative of the cotangent function, so

\[ g'(x) = (1)(\cot{x})+x(-\csc^2{x}),\]

and finally, simplify the derivative, that is

\[ g'(x) = \cot{x}-x\csc^2{x}.\]

Simple, right? Take a look at one more example using the Chain Rule.

Find the derivative of

\[ h(x)= e^{\csc{x}}.\]

**Solution:**

Here you need to identify that the exponential function takes the output of the cosecant function as input, this means that you should let

\[ u = \csc{x},\]

so, using the derivative of the cosecant function you get

\[ \frac{\mathrm{d}u}{\mathrm{d}x} = -\csc{x}\,\cot{x}.\]

Using the Chain Rule you obtain

\[ h'(x) = \left(\frac{\mathrm{d}}{\mathrm{d}u}e^u\right) \frac{\mathrm{d}u}{\mathrm{d}x}.\]

The derivative of the exponential function is the exponential function itself, so

\[ h'(x) = e^u \frac{\mathrm{d}u}{\mathrm{d}x}.\]

Finally, substitute \(u\) and its derivative and rearrange, so

\[ \begin{align} h'(x) &= (e^{\csc{x}})(-\csc{x}\,\cot{x}) \\ &= -e^{\csc{x}}\csc{x}\,\cot{x}. \end{align}\]

## Derivatives of sec, csc, and cot - Key takeaways

- The secant, cosecant, and cotangent functions are collectively known as the
**reciprocal**trigonometric functions.- The secant function is the reciprocal of the cosine function, \[\sec{x}=\frac{1}{\cos{x}}.\]
- The cosecant function is the reciprocal of the sine function, \[\csc{x}=\frac{1}{\sin{x}}.\]
- The cotangent function is the reciprocal of the tangent function, \[\cot{x}=\frac{1}{\tan{x}}.\]

- You can find the derivatives of the secant, cosecant, and cotangent function using the derivatives of the sine and cosine functions, along with the Quotient Rule.
- \[ \frac{\mathrm{d}}{\mathrm{d}x}\sec{x} = \sec{x}\,\tan{x}.\]
- \[ \frac{\mathrm{d}}{\mathrm{d}x}\csc{x} = -\csc{x}\,\cot{x}.\]
- \[ \frac{\mathrm{d}}{\mathrm{d}x}\cot{x} = -\csc^2{x}.\]

- The inverse trigonometric functions, also known as arcus functions, are the inverse functions of the trigonometric functions. The inverse trigonometric functions are
**not**the same as the reciprocal trigonometric functions. - The derivatives of the inverse trigonometric functions can be obtained using implicit differentiation and some trigonometric identities.
- \[\frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arcsec}{\,x} = \frac{1}{|x|\sqrt{x^2-1}}.\]
- \[ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccot}{\, x} = -\frac{1}{x^2+1}\]
- \[\frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccsc}{\, x} = -\frac{1}{|x|\sqrt{x^2-1}}.\]

###### Learn with 17 Derivatives of Sec, Csc and Cot flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Derivatives of Sec, Csc and Cot

Is csc opposite of sin?

Rather than opposite we say that the cosecant function is the reciprocal of the sine function. This means that csc(x)=1/sin(x).

What is the derivative of sec?

The derivative of the secant function is the secant function times the tangent function.

What is the derivative of csc?

The derivative of the cosecant function is the negative of the cosecant function times the cotangent function.

What is the derivative of cot?

The derivative of the cotangent function is the negative of the square of the cosecant function.

Is cos and cosec same?

No. Usually, "cos" refers to the cosine function, while "csc" refers to the cosecant function.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more