Explainer: Differentiation of Trigonometric Functions

In this explainer, we will learn how to find the derivatives of trigonometric functions and how to apply the differentiation rules on them.

Having learned how to differentiate polynomial functions, we would like to extend our knowledge to the derivatives of trigonometric functions.

We will begin by considering the sine function and what we can learn about its derivative by considering its graph. Recall that we can extend sine to be a function defined on all real numbers π‘₯ and we write 𝑓(π‘₯)=π‘₯,sin where π‘₯ is measured in radians. We will begin by considering the graph of the sine function.

We can now try to sketch a graph of its derivative. Notice that the turning points are at πœ‹2+π‘˜πœ‹ for π‘˜βˆˆβ„€β€”this is where the derivative will be zero. Furthermore, at the point where the sine graph meets the π‘₯-axis, its derivative is at a maximum or minimum. In particular, when π‘₯=2π‘˜πœ‹ for π‘˜βˆˆβ„€, the slope of its tangent line is 1 and when π‘₯=(2π‘˜+1)πœ‹ for π‘˜βˆˆβ„€, the slope of the tangent line is βˆ’1. Hence, the graph of the derivative of sine is as follows.

An astute eye will notice that the graph of the derivative of sine appears to be the same as the graph of the cosine function. This is in fact true, and we will use the next example to prove this fact. However, the technique of considering the graphs of functions and sketching their derivatives can help us gain insight about the nature of its derivate. Before we formally differentiate sine, we recap the definition of the derivative of a function.

Derivative

The derivative of a function 𝑓 is defined as 𝑓(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„ŽοŽ˜ο‚β†’οŠ¦lim at the points where the limit exists.

Example 1: Differentiating Sine from First Principles

Differentiate 𝑓(π‘₯)=π‘₯sin from first principles.

Answer

Recall that the derivative of a function 𝑓 is defined by the limit 𝑓(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„Ž.οŽ˜ο‚β†’οŠ¦lim

Substituting in sine for 𝑓, we have 𝑓(π‘₯)=(π‘₯+β„Ž)βˆ’(π‘₯)β„Ž.οŽ˜ο‚β†’οŠ¦limsinsin

To simplify the expression within the limit, we can use our knowledge of trigonometric identities. In particular, we can use the sum formula, sinsincoscossin(𝐴+𝐡)=𝐴𝐡+𝐴𝐡, to rewrite this expression as 𝑓(π‘₯)=π‘₯β„Ž+π‘₯β„Žβˆ’(π‘₯)β„Ž=π‘₯(β„Žβˆ’1)+π‘₯β„Žβ„Ž.οŽ˜ο‚β†’οŠ¦ο‚β†’οŠ¦limsincoscossinsinlimsincoscossin

Using the sum and product rules of limits, we can rewrite this as 𝑓(π‘₯)=ο€Όπ‘₯οˆο€½β„Žβˆ’1β„Žο‰+ο€Όπ‘₯οˆο€½β„Žβ„Žο‰.οŽ˜ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦limsinlimcoslimcoslimsin

Since both sinπ‘₯ and cosπ‘₯ are independent of β„Ž, their limits will simply be sinπ‘₯ and cosπ‘₯ respectively. Hence,

𝑓(π‘₯)=π‘₯ο€½β„Žβˆ’1β„Žο‰+π‘₯ο€½β„Žβ„Žο‰.οŽ˜ο‚β†’οŠ¦ο‚β†’οŠ¦sinlimcoscoslimsin(1)

The other two limits are not as trivial. However, limsinο‚β†’οŠ¦ο€½β„Žβ„Žο‰=1 is actually a standard result. It is possible to derive this by proving that cossinβ„Ž<β„Žβ„Ž<1 and applying the squeeze theorem. However, we will not give the details of that here. We can also note that sinπ‘₯β‰ˆπ‘₯ for small π‘₯, which also points to the fact that the limit will be equal to one. Notice also that, for this limit to evaluate to one, we require that π‘₯ be measured in radians. For this result, it is not the case if π‘₯ is measured in degrees.

We now turn our attention to the second limit, limcosο‚β†’οŠ¦ο€½β„Žβˆ’1β„Žο‰.

We begin by multiplying the numerator and denominator by the conjugate of the numerator, which yields limcoslimcoscoscoslimcoscosο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦οŠ¨ο€½β„Žβˆ’1β„Žο‰=ο€½β„Žβˆ’1β„Žβ„Ž+1β„Ž+1=ο€Ύβ„Žβˆ’1β„Ž(β„Ž+1).

Using the Pythagorean identity for sine and cosine, we can rewrite the numerator in terms of sine: limcoslimsincoslimsinsincosο‚β†’οŠ¦ο‚β†’οŠ¦οŠ¨ο‚β†’οŠ¦ο€½β„Žβˆ’1β„Žο‰=ο€Ώβˆ’β„Žβ„Ž(β„Ž+1)=βˆ’ο€½β„Žβ„Žβ„Žβ„Ž+1.

Applying the multiplicative law of limits, we have limcoslimsinlimsincosο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦ο€½β„Žβˆ’1β„Žο‰=βˆ’ο€½β„Žβ„Žο‰ο€½β„Žβ„Ž+1.

Taking the limits, we have limcosο‚β†’οŠ¦ο€½β„Žβˆ’1β„Žο‰=1ο€Ό01+1=0.

This is not surprising given that, for small values of π‘₯, cosπ‘₯ can be approximated by cosπ‘₯β‰ˆ1βˆ’12π‘₯.

Substituting the value of these two limits into (1), we get 𝑓(π‘₯)=π‘₯.cos

We will now consider the derivative of the cosine function. We could use a similar technique as we used for deriving the derivative of sine. However, rather than using the sum and difference trigonometric identities, we will use the product to sum formulae to demonstrate an alternative derivation.

Example 2: Differentiating Cosine from First Principles

Given that 𝑦=π‘₯cos, find dd𝑦π‘₯ from first principles.

Answer

Recall that the derivative of a function 𝑦=𝑓(π‘₯) is defined by the limit ddlim𝑦π‘₯=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„Ž.ο‚β†’οŠ¦

Substituting in cosine for 𝑓, we have ddlimcoscos𝑦π‘₯=(π‘₯+β„Ž)βˆ’(π‘₯)β„Ž.ο‚β†’οŠ¦

To simplify the expression within the limit, we can use our knowledge of trigonometric identities. In particular, we can use the product to sum formula, coscossinsinπ΄βˆ’π΅=βˆ’2𝐴+𝐡2οˆο€Όπ΄βˆ’π΅2, to rewrite this expression as ddlimsinsinlimsinsin𝑦π‘₯=1β„Žο€½βˆ’2ο€½2π‘₯+β„Ž2ο‰ο€½β„Ž2=βˆ’ο€½π‘₯+β„Ž2.ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚οŠ¨ο‚οŠ¨

Using the properties of limits, we can rewrite this as ddlimsinlimsin𝑦π‘₯=ο€½βˆ’ο€½π‘₯+β„Ž2.ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚οŠ¨ο‚οŠ¨

The first of these limits simply evaluates to βˆ’π‘₯sin. As for the second, it is equivalent to limsinοΌβ†’οŠ¦πœƒπœƒ. Using standard results, we find this is equal to one. Hence, ddsin𝑦π‘₯=βˆ’π‘₯.

These formulae for the derivatives of sine and cosine can be generalized to the derivatives of sinπ‘Žπ‘₯ and cosπ‘Žπ‘₯ as follows: ddsincosddcossinπ‘₯π‘Žπ‘₯=π‘Žπ‘Žπ‘₯,π‘₯π‘Žπ‘₯=βˆ’π‘Žπ‘Žπ‘₯.

To differentiate the other trigonometric functions, we will appeal to the rules of derivatives, in particular, to the quotient rule. Below is a summary of the product and quotient rules.

Product Rule

Given two differentiable functions 𝑒 and 𝑣, the derivative of their product is given by ddddddπ‘₯(𝑒(π‘₯)𝑣(π‘₯))=𝑒(π‘₯)π‘₯(𝑣(π‘₯))+𝑣(π‘₯)π‘₯(𝑒(π‘₯)).

This can be written succinctly using prime notation as follows: (𝑒𝑣)=𝑒𝑣+𝑒𝑣.

Quotient Rule

Given two differentiable functions 𝑒 and 𝑣, the derivative of their quotient is given by ddπ‘₯𝑒(π‘₯)𝑣(π‘₯)=𝑣(π‘₯)(𝑒(π‘₯))βˆ’π‘’(π‘₯)(𝑣(π‘₯))(𝑣(π‘₯)).ddddο—ο—οŠ¨

This can be written succinctly using prime notation as follows: 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£.

Example 3: Derivative of the Tangent Function

Evaluate the rate of change of 𝑓(π‘₯)=5π‘₯tan at π‘₯=πœ‹.

Answer

Recall that we can find the rate of change of a given function at a given point by finding the value of its derivative at the point. Hence, we will first find the derivative of 𝑓.

Using the trigonometric identity that tansincosπ‘₯=π‘₯π‘₯, we can express 𝑓 as 𝑓(π‘₯)=5π‘₯5π‘₯.sincos

To find the derivative of 𝑓, we can use the quotient rule: 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£, setting 𝑒=5π‘₯sin and 𝑣=5π‘₯cos. We begin by differentiating 𝑒 and 𝑣 to find expressions for π‘’οŽ˜ and π‘£οŽ˜. Starting with 𝑒, we know the derivative of sinπ‘Žπ‘₯ is π‘Žπ‘Žπ‘₯cos. Hence, 𝑒=55π‘₯.cos

Similarly, 𝑣=βˆ’55π‘₯sin

Substituting these expressions into the quotient rule, we have 𝑓(π‘₯)=5π‘₯(55π‘₯)βˆ’(βˆ’55π‘₯)5π‘₯(5π‘₯)=5ο€Ί5π‘₯+5π‘₯5π‘₯.coscossinsincoscossincos

Using the Pythagorean identity for sine and cosine, we can simplify the numerator. Hence, 𝑓(π‘₯)=55π‘₯.=55π‘₯.cossec

We can now substitute in π‘₯=πœ‹ to find the rate of change of 𝑓 at this point which gives 𝑓(πœ‹)=55πœ‹=5.sec

The previous example demonstrated that ddtansecπ‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯.

Below is a graph of the tangent function and its derivative.

Thus far, we have looked at the derivatives of sine, cosine, and tangent. Below is a summary of their derivatives.

Derivatives of Trigonometric Functions

The derivatives of the trigonometric functions are as follows: ddsincosddcossinddtansecπ‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=βˆ’π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯.

Generally, you will be able to quote and use these results without derivation. However, you might be expected to derive these results using the techniques we have outlined above. Furthermore, it is certainly helpful to commit these to memory. This will dramatically increase your ability to tackle differentiation problems with ease and efficiency.

We will now consider a number of examples where we can apply these results without derivation.

Example 4: Consecutive Derivatives of Sine

Find the thirty-third derivative of 𝑓(π‘₯)=π‘₯sin.

Answer

It would certainly be tedious to differentiate this function thirty-three times in a row. Therefore, we will begin by looking at the first few derivatives to see whether there is a pattern. The derivative of the sine function is the cosine function. Hence, 𝑓(π‘₯)=π‘₯.cos

The cosine function differentiates to the negative sine function: 𝑓=βˆ’π‘₯.οŽ™sin

Continuing, we find 𝑓=βˆ’π‘₯,𝑓=π‘₯,οŽ›(οŠͺ)cossin which is equal to the function we started with. Hence, this pattern will repeat for higher-order derivatives. Therefore, the general rule is 𝑓=π‘₯,𝑓=π‘₯,𝑓=βˆ’π‘₯,𝑓=βˆ’π‘₯,(οŠͺ)(οŠͺο‡οŠ°οŠ§)(οŠͺο‡οŠ°οŠ¨)(οŠͺο‡οŠ°οŠ©)sincossincos for π‘˜βˆˆβ„€. Since 33=4Γ—8+1, we have 𝑓=π‘₯.()cos

Using a similar argument where we set 𝑓(π‘₯)=π‘₯cos, we can derive the general formula 𝑓=π‘₯,𝑓=βˆ’π‘₯,𝑓=βˆ’π‘₯,𝑓=π‘₯,(οŠͺ)(οŠͺο‡οŠ°οŠ§)(οŠͺο‡οŠ°οŠ¨)(οŠͺο‡οŠ°οŠ©)cossincossin where π‘˜βˆˆβ„€, for the higher-order derivatives of the cosine function.

Example 5: Using the Product Rule with Trigonometric Functions

If 𝑦=π‘₯5π‘₯sin, determine dd𝑦π‘₯.

Answer

This function is a product of two differentiable functions 𝑒=π‘₯ and 𝑣=5π‘₯sin. We can, therefore, use the product rule which states that, for a function 𝑦=𝑒𝑣, dddddd𝑦π‘₯=𝑒𝑣π‘₯+𝑣𝑒π‘₯ to find its derivative. We begin by differentiating 𝑒 and 𝑣. Using the rule for differentiating monomials, we have dd𝑒π‘₯=5π‘₯.οŠͺ

Similarly, using the rule for differentiating sine, ddsincosπ‘₯π‘Žπ‘₯=π‘Žπ‘Žπ‘₯, we have ddcos𝑣π‘₯=55π‘₯.

We can now substitute these expressions into the formula for the product rule as follows: ddcossin𝑦π‘₯=5π‘₯5π‘₯+5π‘₯5π‘₯.οŠͺ

Example 6: Differentiating Trigonometric Functions

If 𝑦=(27π‘₯+27π‘₯)sincos, find dd𝑦π‘₯.

Answer

When faced with a question like this, we might think about applying the product rule or the chain rule for composite functions. However, often times it is useful to think whether we can simplify the expression using trigonometric identities that we already know. This is the approach we will take to demonstrate how this significantly simplifies finding the derivative. We begin by factoring the 2 out of the parentheses: 𝑦=2(7π‘₯+7π‘₯).sincos

We can now expand the parentheses as follows: 𝑦=4ο€Ί7π‘₯+27π‘₯7π‘₯+7π‘₯.sinsincoscos

We can now use the Pythagorean identity for sine and cosine to simplify the expression as follows: 𝑦=4(1+27π‘₯7π‘₯).sincos

We can now use the double angle formula, sinsincos2𝐴=2𝐴𝐴, to rewrite this as 𝑦=4(1+14π‘₯)=4+414π‘₯.sinsin

We now have a much simpler expression for the function which we can easily differentiate using the rules of differentiation for trigonometric functions. Using the fact that ddsincosπ‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯ and that the derivative of a constant term is zero, we have ddcos𝑦π‘₯=5614π‘₯.

The previous example demonstrates an important point regarding differentiation in general but is particularly poignant for trigonometric functions: we can often use the knowledge we have to manipulate the expression for the function to simplify the process of taking the derivative. We will finish by looking at one last example where we apply the quotient rule to evaluate the derivative.

Example 7: Using the Quotient Rule with Trigonometric Functions

If 𝑦=6π‘₯1βˆ’6π‘₯cossin, find dd𝑦π‘₯.

Answer

The function we have been given is a quotient of two differential functions. Therefore, we can apply the quotient rule which states that, for a function 𝑦=𝑒𝑣, the derivative dd𝑦π‘₯=π‘£βˆ’π‘’π‘£.ddddο‘ο—ο“ο—οŠ¨

Setting 𝑒=6π‘₯cos and 𝑣=1βˆ’6π‘₯sin, we begin by finding the derivatives of 𝑒 and 𝑣. Using the rule of differentiation for trigonometric functions, ddsincosddcossinπ‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=βˆ’π‘Žπ‘Žπ‘₯, we have ddsin𝑒π‘₯=βˆ’66π‘₯ and ddcos𝑣π‘₯=βˆ’66π‘₯.

Substituting these expressions back into the quotient rule, we have ddsinsincoscossinsinsincossin𝑦π‘₯=(1βˆ’6π‘₯)(βˆ’66π‘₯)βˆ’(6π‘₯)(βˆ’66π‘₯)(1βˆ’6π‘₯)=βˆ’66π‘₯+66π‘₯+66π‘₯(1βˆ’6π‘₯).

Using the Pythagorean identity for sine and cosine, we can simplify the numerator as follows: ddsinsin𝑦π‘₯=6(1βˆ’6π‘₯)(1βˆ’6π‘₯).

Finally, we cancel the common factor to get ddsin𝑦π‘₯=61βˆ’6π‘₯.

Key Points

  1. We can extend our understanding of derivatives to trigonometric functions.
  2. The derivatives of the sine, cosine, and tangent functions are as follows: ddsincosddcossinddtansecπ‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=βˆ’π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯.
  3. The higher-order derivatives of both the sine and the cosine functions form a repeating pattern. For sine, the pattern is as follows: ddsinsinddsincosddsinsinddsincos(οŠͺ)(οŠͺ)(οŠͺο‡οŠ°οŠ§)(οŠͺο‡οŠ°οŠ§)(οŠͺο‡οŠ°οŠ¨)(οŠͺο‡οŠ°οŠ¨)(οŠͺο‡οŠ°οŠ©)(οŠͺο‡οŠ°οŠ©)π‘₯(π‘₯)=π‘₯,π‘₯(π‘₯)=π‘₯,π‘₯(π‘₯)=βˆ’π‘₯,π‘₯(π‘₯)=βˆ’π‘₯. For cosine, the pattern is ddcoscosddcossinddcoscosddcossin(οŠͺ)(οŠͺ)(οŠͺο‡οŠ°οŠ§)(οŠͺο‡οŠ°οŠ§)(οŠͺο‡οŠ°οŠ¨)(οŠͺο‡οŠ°οŠ¨)(οŠͺο‡οŠ°οŠ©)(οŠͺο‡οŠ°οŠ©)π‘₯(π‘₯)=π‘₯,π‘₯(π‘₯)=βˆ’π‘₯,π‘₯(π‘₯)=βˆ’π‘₯,π‘₯(π‘₯)=π‘₯.
  4. Using these standard results, we can find the derivatives of a large class of functions.

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