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# Derivative of Trigonometric Functions

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Let's take a break and think of the beach for a moment. We can see the waves in the sea, a volleyball bouncing up and down. If we focus on the horizon, we can see a buoy floating. It goes up and down along the sea waves!

A common scenario present at a beach, pixabay.com

What do all these things have in common? The answer is that their movement is periodic. Periodic functions describe things like sea waves. Periodic functions are functions that repeat their outputs at regular intervals. Trigonometric functions are prime examples of periodic functions. For this reason, it is essential to know how to differentiate trigonometric functions.

## The Meaning of the Derivative of Trigonometric Functions

You might be wondering what does it mean to find the derivative of a trigonometric function.

Finding the derivative of a function means that you are finding another function that describes its rate of change.

That is, the derivative of a function is another function which describes how the original function changes. This is done independently of which kind of functions you are dealing with, and trigonometric functions are no exception!

Formulas are usually given for the derivatives of all kinds of functions. Here you will find how to find the derivatives of trigonometric functions.

## Formulas for the Derivatives of Trigonometric Functions

There are six main trigonometric functions:

• The Sine Function: $$\sin{x}.$$

• The Cosine Function: $$\cos{x}.$$

• The Tangent Function: $$\tan{x}.$$

• The Cotangent Function: $$\cot{x}.$$

• The Secant Function: $$\sec{x}.$$

• The Cosecant Function: $$\csc{x}.$$

Trigonometric functions are the bridge between trigonometry and calculus. All six trigonometric functions are periodic functions.

For a reminder about the graphs of these functions and their periods, see Trigonometric Functions.

Let's now take a look at each one of their derivatives.

The derivatives of the main trigonometric functions are:

$$\frac{\mathrm{d}}{\mathrm{d}x}\sin{x}=\cos{x},$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}\tan{x}=\sec^2{x},$$

$$\frac{\mathrm{d}}{\mathrm{d}x}\cot{x}=-\csc^2{x},$$

$$\frac{\mathrm{d}}{\mathrm{d}x}\sec{x}=\left( \sec{x} \right)\left(\tan{x}\right),$$

and

$$\frac{\mathrm{d}}{\mathrm{d}x}\csc{x}=-\left( \csc{x} \right)\left(\cot{x}\right).$$

Note how all the derivatives of the trigonometric functions involve more trigonometric functions. This connection is a signature of the periodicity of trigonometric functions!

Let's see how to find the derivative of some trigonometric functions using the above derivatives along with basic Differentiation Rules.

## Derivatives of Trigonometric Functions and the Chain Rule

Let's take a look at how to differentiate trigonometric functions using the Chain Rule.

Find the derivative of $$f(x)=\sin{2x}.$$

To find this derivative you will need to use the Chain Rule. Let $$u=2x.$$ Then by the Power Rule,

$$u'(x)=2.$$

So now using the Chain Rule,

\begin{align}f'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}u}\sin{u} \right) \left( \frac{\mathrm{d}u}{\mathrm{d}x} \right) \0.5em] &= \left( \cos{u} \right) (2) \\ &= 2\cos{u}. \end{align} Finally, substitute back $$u=2x,$$ so f'(x)=2\cos{2x}. Do not forget to square the secant function when differentiating the tangent function! Find the derivative of $$g(x)=\tan{x^3}.$$ Answer: Start by letting $$u=x^3.$$ By the Power Rule, u'(x)=3x^2. Next, use the Chain Rule, \begin{align}g'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}u}\tan{u} \right) \left( \frac{\mathrm{d}u}{\mathrm{d}x} \right) \\[0.5em] &= \left(\sec^2{u} \right) (3x^2) \\ &= 3x^2\sec^2{u}, \end{align} and substitute back $$u=x^3,$$ obtaining g'(x)=3x^2\sec^2{x^3}. Remember that you have two functions for the derivatives of the secant and cosecant functions. Do not forget about either when substituting back $$u.$$ Find the derivative of $$h(x)=\csc{2x^2}.$$ Answer: Start by letting $$u=2x^2.$$ By the Power Rule, u'(x)=4x. Next, use the Chain Rule to get \begin{align}h'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}u}\csc{u} \right) \left( \frac{\mathrm{d}u}{\mathrm{d}x} \right) \\[0.5em] &= \left(-\csc{u} \right) \left( \cot{u} \right) (4x) \\ &= -4x\left(\csc{u}\right) \left(\cot{u} \right) , \end{align} and substitute back $$u=2x^2$$ to get h'(x)=-4x\left(\csc{2x^2}\right) \left(\cot{2x^2} \right). ## Examples of Derivatives of Trigonometric Functions Calculus is all about practice! You also need to be able to use more differentiation rules, like the Product Rule and the Quotient Rule. Find the derivative of $$f(x)=x \left(\sin{x}\right).$$ Answer: Since you have a product of functions start by using the Product Rule, that is f'(x)=\left( \frac{\mathrm{d}}{\mathrm{d}x} x \right) \sin{x} + x \left(\frac{\mathrm{d}}{\mathrm{d}x} \sin{x} \right). You can find the derivative of $$x$$ by using the Power Rule, \frac{\mathrm{d}}{\mathrm{d}x} x= 1, and the derivative of the sine function is the cosine function \frac{\mathrm{d}}{\mathrm{d}x}\sin{x}=\cos{x}. Knowing this, the derivative of $$f(x)$$ is \begin{align}f'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}x} x \right) \sin{x} + x \left(\frac{\mathrm{d}}{\mathrm{d}x} \sin{x} \right) \\[0.5em] &= (1)\sin{x}+x\left( \cos{x} \right) \\ &= \sin{x}+x \cos{x}.\end{align} How about The Quotient Rule? No problem! Find the derivative of  g(x) = \frac{\tan{x}}{x^2}. Answer: Now you have a quotient of functions, so start by using the Quotient Rule, that is g'(x)=\frac{ \left( \frac{\mathrm{d}}{\mathrm{d}x} \tan{x} \right)x^2-\tan{x}\left( \frac{\mathrm{d}}{\mathrm{d}x} x^2 \right) }{\left( x^2 \right)^2}. You can find the derivative of $$x^2$$ with the Power Rule, \frac{\mathrm{d}}{\mathrm{d}x} x^2=2x, and the derivative of the tangent function is the secant function squared \frac{\mathrm{d}}{\mathrm{d}x} \tan{x}=\sec^2{x}. Finally, substitute the above derivatives in the Quotient Rule and simplify, obtaining \begin{align}g'(x) &= \frac{ \left( \sec^2{x} \right) x^2- \left( \tan{x} \right) (2x) }{ \left( x^2 \right) ^2} \\[0.5em] &= \frac{x^2 \left( \sec^2{x} \right) -2x \left( \tan{x} \right) }{x^4} \\[0.5em] &= \frac{x \left( \sec^2{x} \right) - 2\tan{x}}{x^3} .\end{align} It's time for one more example using the Chain Rule. Find the derivative of $$h(x)=\sin^2{x}.$$ Answer: Since the sine function is squared, you are dealing with a composition of functions, hence you need to use the Chain Rule. Start by letting $$u=\sin{x}.$$ Its, derivative is the cosine function u'(x)=\cos{x}. Next, use the Chain Rule, \begin{align}h'(x) &= \left( \frac{\mathrm{d}}{\mathrm{d}u}u^2 \right) \left( \frac{\mathrm{d}u}{\mathrm{d}x} \right) \\[0.5em] &= \left(2u \right) \left( \cos{x} \right) \\ &= 2u \left(\cos{u} \right), \end{align} where you have used the Power Rule to find the derivative of $$u^2.$$ Finally, substitute back $$u=\sin{x},$$ obtaining h'(x)=2 \left( \sin{x} \right) \left( \cos{x} \right) . Remember, practice makes perfect! ## Common Mistakes Everyone makes mistakes from time to time. Here you will see some common mistakes when differentiating trigonometric functions. One common mistake is getting the signs mistaken when differentiating the cosine function, the cotangent function, or the cosecant function, that is \frac{\mathrm{d}}{\mathrm{d}x} \cos{x} \neq \sin{x}.  Be sure to put the negative sign! \frac{\mathrm{d}}{\mathrm{d}x} \cos{x} = -\sin{x}.  One easy way to remember the signs of the derivatives of the trigonometric functions is to pay attention to the name of the function. If it starts with “co”, like cosine, cotangent, and cosecant, then the derivative has a negative sign. Another common mistake happens when differentiating the secant function or the cosecant function. Remember that when differentiating these functions you have to write the correct inputs in all instances of trigonometric functions. \frac{\mathrm{d}}{\mathrm{d}x} \sec{x^2} \neq 2x \left(\sec{x^2} \right) \left( \tan{x} \right).  Here, the square of the input of the tangent function is missing. Find the derivative using the derivative of the secant function formula. Do not forget to use any relevant differentiation techniques, like the Chain Rule in this case. \frac{\mathrm{d}}{\mathrm{d}x} \sec{x^2} = 2x \left(\sec{x^2} \right) \left( \tan{x^2} \right).  Be careful if you are differentiating trigonometric functions with different inputs. Doing things step-by-step will help you not get the inputs mixed up! ## Derivatives of Inverse Trigonometric Functions You might also need to find the derivatives of the inverse trigonometric functions, like the inverse sine, the inverse tangent, and so on. Check out our article on Derivatives of Inverse Trigonometric Functions for a deep discussion! ## Derivatives of Trigonometric Functions - Key takeaways • Trigonometric functions are periodic functions. Trigonometric functions are used to describe periodic phenomena. • The derivatives of the six main trigonometric functions are the following:\frac{\mathrm{d}}{\mathrm{d}x}\sin{x}=\cos{x}, \frac{\mathrm{d}}{\mathrm{d}x}\cos{x}=-\sin{x}, \frac{\mathrm{d}}{\mathrm{d}x}\tan{x}=\sec^2{x}, \frac{\mathrm{d}}{\mathrm{d}x}\cot{x}=-\csc^2{x}, \frac{\mathrm{d}}{\mathrm{d}x}\sec{x}=\left( \sec{x} \right)\left(\tan{x}\right), and \frac{\mathrm{d}}{\mathrm{d}x}\csc{x}=-\left( \csc{x} \right)\left(\cot{x}\right). • Common mistakes when differentiating trigonometric functions include the following: • Getting the signs mistaken. Remember that functions that start with “co” have a negative sign in their derivative. • Mixing the inputs of the derivatives of the secant function and the cosecant function. Remember to place the correct input in each trigonometric function after differentiating. ## Frequently Asked Questions about Derivative of Trigonometric Functions The derivatives of trigonometric functions are the following: • The derivative of the sine function is the cosine function. • The derivative of the cosine function is the negative sine function. The derivatives of the rest of the trigonometric functions can be found using the quotient rule and trigonometric identities. Trigonometric functions are functions that take inputs as angles in radians and assign them to the corresponding trigonometric ratio. One example of the derivative of trigonometric functions is that the derivative of the sine function is the cosine function. You should start by inspecting the function to see if any relevant differentiation technique is needed, like the chain rule or the product rule. You can then find the derivatives of the trigonometric functions, which are usually given in derivatives tables. It depends on the trigonometric function you want to take the derivative of, but in general you can use the definition of the derivative and take the limit, just like with any other function. ## Final Derivative of Trigonometric Functions Quiz Question Apart from using the definition of a derivative, how can you prove the derivative of the tangent function? Show answer Answer Using the quotient rule and the derivatives of sine and cosine functions. Show question Question What are the steps for finding the derivative of an inverse trigonometric function? Show answer Answer 1. Identify which differentiation rule(s) is (are) relevant. 2. Use the above differentiation rules. 3. Write the derivative(s) of the inverse trigonometric function(s), as well as any other functions involved in the calculation. Show question Question The derivatives of the six main trigonometric functions are all _______. Show answer Answer Trigonometric functions Show question Question Trigonometric functions are used to describe _______ behavior. Show answer Answer periodic Show question Question The derivatives of these functions involve a negative sign. Show answer Answer Cosine Show question Question The derivatives of these functions involve the product of two different trigonometric functions. Show answer Answer Secant Show question Question The derivatives of these functions involve the square of another trigonometric function. Show answer Answer Tangent Show question Question The secant, cosecant, and cotangent functions are collectively known as the ____ functions. Show answer Answer reciprocal trigonometric. Show question Question The secant function is the reciprocal of the ____ function. Show answer Answer cosine. Show question Question The cosecant function is the reciprocal of the ____ function. Show answer Answer sine. Show question Question The cotangent function is the reciprocal of the ____ function. Show answer Answer tangent. Show question Question Which of the following expressions is the derivative of the secant function? Show answer Answer \[\frac{\mathrm{d}}{\mathrm{d}x} \sec{x} = \sec{x}\,\tan{x}.

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Question

Which of the following expressions is the derivative of the cosecant function?

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

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Question

Which of the following expressions is the derivative of the cotangent function?

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

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Question

True/False: Inverse trigonometric functions are the same as reciprocal trigonometric functions.

False.

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Question

The expression

$\sec^{-1}{x}$

corresponds to ____.

the inverse secant function.

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Question

The expression

$\cot^{-1}{x}$

corresponds to ____.

the inverse cotangent function.

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Question

This function is the reciprocal of the sine function.

The cosecant function.

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Question

This function is the reciprocal of the cosine function.

The secant function.

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Question

This function is the reciprocal of the tangent function.

The cotangent function.

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Question

The cotangent function can be written in terms of the sine and the cosine functions as:

$\cot{x}=\frac{\cos{x}}{\sin{x}}$.

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Question

Which of the following expressions is the derivative of the inverse secant function?

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

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Question

Which of the following expressions is the derivative of the inverse cotangent function?

$\frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccot}{\, x} =-\frac{1}{x^2+1}$.

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Question

Which of the following expressions is the derivative of the inverse cosecant function?

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

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