Lesson Explainer: Differentiation of Trigonometric Functions | Nagwa Lesson Explainer: Differentiation of Trigonometric Functions | Nagwa

Lesson Explainer: Differentiation of Trigonometric Functions Mathematics • Second Year of Secondary School

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

Trigonometric functions and their derivatives have several real-world applications in various fields such as physics, engineering, architecture, robotics, music theory, and navigation, to name a few. In physics, it can be used in projectile motion, to model the mechanics of electromagnetic waves, analyzing alternating and direct currents, and finding the trajectory of a mass around a massive body under the force of gravity.

In this explainer, we will particularly be interested in the derivatives of the sine, cosine, and tangent functions. We will first determine the derivative of the standard trigonometric functions, starting with sinπ‘₯, from first principles, and then use that result to also determine the derivatives of cosπ‘₯ and tanπ‘₯ using the chain and quotient rule.

Let’s first recall the definition of the derivative.

Definition: The Derivative

The derivative of a function 𝑓(π‘₯) is defined as 𝑓′(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„Žο‰,limο‚β†’οŠ¦ at the points where the limit exists.

Substituting 𝑓(π‘₯)=π‘₯sin into the definition of the derivative, we have 𝑓′(π‘₯)=ο€½(π‘₯+β„Ž)βˆ’π‘₯β„Žο‰.limsinsinο‚β†’οŠ¦

To simplify the expression within the limit, we will make use of trigonometric identities. In particular, we can use the sum identity, sinsincoscossin(𝐴+𝐡)=𝐴𝐡+𝐴𝐡.

Using this identity, we can expand the sin(π‘₯+β„Ž) and rewrite the resulting expression using the sum and product rules of limits to obtain 𝑓′(π‘₯)=ο€½π‘₯β„Ž+π‘₯β„Žβˆ’π‘₯β„Žο‰=ο€½π‘₯β„Ž+π‘₯(β„Žβˆ’1)β„Žο‰=ο€½π‘₯Γ—β„Žβ„Ž+π‘₯Γ—β„Žβˆ’1β„Žο‰=ο€Όπ‘₯οˆο€½β„Žβ„Žο‰+ο€Όπ‘₯οˆο€½β„Žβˆ’1β„Žο‰.limsincoscossinsinlimcossinsincoslimcossinsincoslimcoslimsinlimsinlimcosο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦

Since both sinπ‘₯ and cosπ‘₯ are independent of β„Ž, their limits will simply be sinπ‘₯ and cosπ‘₯ respectively. Hence, the derivative becomes 𝑓′(π‘₯)=π‘₯ο€½β„Žβ„Žο‰+π‘₯ο€½β„Žβˆ’1β„Žο‰.coslimsinsinlimcosο‚β†’οŠ¦ο‚β†’οŠ¦

The other two limits are not as trivial but are standard results:

  1. limsinο‚β†’οŠ¦ο€½β„Žβ„Žο‰=1;
  2. limcosο‚β†’οŠ¦ο€½β„Žβˆ’1β„Žο‰=0.

We can also determine the second limit from the first by applying trigonometric identities. In order to see this, we begin by multiplying the numerator and denominator by cosβ„Ž+1, which yields limcoslimcoscoscoslimcoscosο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦οŠ¨ο€½β„Žβˆ’1β„Žο‰=ο€½β„Žβˆ’1β„ŽΓ—β„Ž+1β„Ž+1=ο€Ύβ„Žβˆ’1β„Ž(β„Ž+1).

Using the Pythagorean identity for sine and cosine, cossinοŠ¨οŠ¨β„Žβˆ’1=βˆ’β„Ž, we can rewrite the numerator in terms of sine and then apply the multiplicative law and take the resulting limits: limcoslimsincoslimsinsincoslimsinlimsincosο‚β†’οŠ¦ο‚β†’οŠ¦οŠ¨ο‚β†’οŠ¦ο‚β†’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=βˆ’ο€½β„Žβ„Žο‰ο€½β„Žβ„Ž+1=βˆ’1ο€Ό01+1=0.

Substituting the value of these two limits into 𝑓′(π‘₯), we obtain the derivative of 𝑓(π‘₯)=π‘₯sin as follows: 𝑓′(π‘₯)=π‘₯ο€½β„Žβ„Žο‰+π‘₯ο€½β„Žβˆ’1β„Žο‰=π‘₯Γ—1+π‘₯Γ—0=π‘₯.coslimsinsinlimcoscossincosο‚β†’οŠ¦ο‚β†’οŠ¦

We can also determine the derivative of 𝑓(π‘₯)=π‘₯cos from first principles in a similar way; however, it is better to determine the result from the cofunction identity cossinπ‘₯=ο€»πœ‹2βˆ’π‘₯ and the chain rule with 𝑓(π‘₯)=ο€»πœ‹2βˆ’π‘₯sin.

Rule: The Chain Rule

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

This can be written succinctly using prime notation as follows: (𝑒(𝑣))β€²=𝑒′(𝑣)𝑣′.

We can rewrite the function 𝑓(π‘₯)=π‘₯cos as 𝑓=𝑒sin with 𝑒=πœ‹2βˆ’π‘₯, and the derivatives are ddcosdd𝑓𝑒=𝑒,𝑒π‘₯=βˆ’1.

Thus, using the chain rule, the derivative of 𝑓(π‘₯)=π‘₯cos is 𝑓′(π‘₯)=𝑓𝑒𝑒π‘₯=βˆ’ο€»πœ‹2βˆ’π‘₯.ddddcos

We can rewrite the final result using the other cofunction identity, cossinο€»πœ‹2βˆ’π‘₯=π‘₯, in reverse: 𝑓′(π‘₯)=βˆ’π‘₯.sin

Now, let’s consider higher derivatives of sinπ‘₯ and cosπ‘₯ that will form a cyclic pattern in the order of the derivatives. For 𝑓(π‘₯)=π‘₯sin, the first four derivatives are ddcosddsinddcosddsin𝑓π‘₯=π‘₯,𝑓π‘₯=βˆ’π‘₯,𝑓π‘₯=βˆ’π‘₯,𝑓π‘₯=π‘₯.οŠͺοŠͺ

Thus, the fourth derivative gives you back the original function, and higher derivatives of sine form a repeating cyclic pattern with period 4. This can be depicted in the following diagram.

The general derivatives for π‘˜βˆˆβ„€ can be written as ddsinformodcosformodsinformodcosformod()()𝑓π‘₯=⎧βŽͺ⎨βŽͺ⎩π‘₯,π‘˜β‰‘0(4),π‘₯,π‘˜β‰‘1(4),βˆ’π‘₯,π‘˜β‰‘2(4),βˆ’π‘₯,π‘˜β‰‘3(4).

Similarly, for 𝑓(π‘₯)=π‘₯cos, the first four derivatives are ddsinddcosddsinddcos𝑓π‘₯=βˆ’π‘₯,𝑓π‘₯=βˆ’π‘₯,𝑓π‘₯=π‘₯,𝑓π‘₯=π‘₯.οŠͺοŠͺ

Thus, the fourth derivative gives you back the original function and higher derivatives of cosine form a repeating cyclic pattern with period 4, similar to the sine function. This can be depicted in the following diagram.

The general derivatives for π‘˜βˆˆβ„€ can be written as ddcosformodsinformodcosformodsinformod()()𝑓π‘₯=⎧βŽͺ⎨βŽͺ⎩π‘₯,π‘˜β‰‘0(4),βˆ’π‘₯,π‘˜β‰‘1(4),βˆ’π‘₯,π‘˜β‰‘2(4),π‘₯,π‘˜β‰‘3(4).

Now, let’s consider an example where we determine a particular higher-order derivative of sine by using the cyclic pattern.

Example 1: Consecutive Derivatives of Sine

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

Answer

In this example, we want to determine the thirty-third derivative of the sine function.

For 𝑓(π‘₯)=π‘₯sin, the first four derivatives are ddcosddsinddcosddsin𝑓π‘₯=π‘₯,𝑓π‘₯=βˆ’π‘₯,𝑓π‘₯=βˆ’π‘₯,𝑓π‘₯=π‘₯.οŠͺοŠͺ

Thus, the fourth derivative gives you back the original function and this cyclic pattern will repeat for higher-order derivatives with period 4. The general rule for the π‘˜th derivative with π‘˜βˆˆβ„€ is ddsinformodcosformodsinformodcosformod()()𝑓π‘₯=⎧βŽͺ⎨βŽͺ⎩π‘₯,π‘˜β‰‘0(4),π‘₯,π‘˜β‰‘1(4),βˆ’π‘₯,π‘˜β‰‘2(4),βˆ’π‘₯,π‘˜β‰‘3(4).

Since 33=4Γ—8+1, we have 33≑1(4)mod. Hence, the thirty-third derivative is equivalent to the first derivative of 𝑓(π‘₯)=π‘₯sin and we have ddddcos()()𝑓π‘₯=𝑓π‘₯=π‘₯.

The derivative of 𝑓(π‘₯)=π‘₯tan can be found by using the quotient rule; by using the trigonometric identity that tansincosπ‘₯=π‘₯π‘₯, we can express 𝑓(π‘₯) as 𝑓(π‘₯)=π‘₯π‘₯.sincos

To find the derivative of 𝑓, we can use the quotient rule.

Rule: The Quotient Rule

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

This can be written succinctly using prime notation as follows: 𝑒𝑣=π‘’β€²π‘£βˆ’π‘’π‘£β€²π‘£.

Setting 𝑒=π‘₯sin and 𝑣=π‘₯cos, we begin by differentiating 𝑒 and 𝑣 to find expressions for 𝑒′ and 𝑣′ from previously established results: 𝑒′=π‘₯,𝑣′=βˆ’π‘₯.cossin

Substituting these expressions into the quotient rule, we have 𝑓′(π‘₯)=π‘₯(π‘₯)βˆ’(βˆ’π‘₯)π‘₯(π‘₯)=ο€Ίπ‘₯+π‘₯π‘₯.coscossinsincoscossincos

Using the Pythagorean identity for sine and cosine, we can simplify the numerator. Hence, 𝑓′(π‘₯)=1π‘₯=π‘₯.cossec

In general, we can establish a rule for derivatives of the function 𝑓(π‘₯)=(π‘Žπ‘₯)sin using the chain rule; we can rewrite this as 𝑓=𝑒sin, where 𝑒=π‘Žπ‘₯, and using ddcosdd𝑓𝑒=𝑒,𝑒π‘₯=π‘Ž the derivative becomes 𝑓′(π‘₯)=𝑓𝑒𝑒π‘₯=π‘Ž(π‘Žπ‘₯).ddddcos

Similarly, for 𝑓(π‘₯)=(π‘Žπ‘₯)cos, using the chain rule by rewriting the function as 𝑓=𝑒cos, where 𝑒=π‘Žπ‘₯, we obtain the general derivative 𝑓′(π‘₯)=βˆ’π‘Ž(π‘Žπ‘₯).sin

The derivative of 𝑓(π‘₯)=(π‘Žπ‘₯)tan can also be found using the chain rule by rewriting the function as 𝑓=𝑒tan, where 𝑒=π‘Žπ‘₯, or using the quotient rule with the derivatives of the general sine and cosine functions to obtain 𝑓′(π‘₯)=π‘Ž(π‘Žπ‘₯).sec

Now, we summarize the general derivatives of trigonometric functions as established.

Standard Result: Derivatives of Trigonometric Functions

The general derivatives of the sine, cosine, and tangent functions are as follows: ddsincosddcossinddtansecπ‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯),π‘₯((π‘Žπ‘₯))=βˆ’π‘Ž(π‘Žπ‘₯),π‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯), for π‘Žβˆˆβ„.

Using these standard results, we can find the derivatives of a large class of functions containing sums of trigonometric functions using the linearity of the derivative: ddddddforπ‘₯(π‘Žπ‘“+𝑏𝑔)=π‘Žπ‘“π‘₯+𝑏𝑔π‘₯,π‘Ž,π‘βˆˆβ„.

Other functions that have compositions, products, or quotients of trigonometric functions use the chain rule, the product rule, the quotient rule, or any combination of them.

In the next example, we will differentiate a function containing a sum of polynomial and trigonometric functions using the linearity of the derivative.

Example 2: Differentiating a Combination of Polynomial and Trigonometric Functions

If 𝑦=βˆ’2π‘₯+6ο€»π‘₯2+ο€»πœ‹4οŠͺsincos, find dd𝑦π‘₯.

Answer

In this example, we want to differentiate a function that has a combination of a polynomial and trigonometric functions.

The power rule and rules for differentiating the sine are ddddsincosπ‘₯(π‘₯)=𝑛π‘₯,π‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯).

Applying these rules, we can calculate the first derivative of 𝑦=βˆ’2π‘₯+6ο€»π‘₯2+ο€»πœ‹4οŠͺsincos as ddddddsinddcoscoscos𝑦π‘₯=βˆ’2π‘₯ο€Ήπ‘₯+6π‘₯ο€»ο€»π‘₯2+π‘₯ο€»ο€»πœ‹4=βˆ’2Γ—4π‘₯+6Γ—12ο€»π‘₯2+0=βˆ’8π‘₯+3ο€»π‘₯2.οŠͺ

In the next example, we will use the chain rule to find the derivative of a function composed of a quadratic function and the sum of sine and cosine functions.

Example 3: Differentiating Trigonometric Functions

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

Answer

In this example, we want to determine the derivative of a composite function using the chain rule and the general rule for differentiating the sine and cosine functions.

Recall that the chain rule for a composite function 𝑦=𝑒(𝑣(π‘₯)) is given by dddddd𝑦π‘₯=𝑒𝑣𝑣π‘₯, for differentiable functions 𝑒 and 𝑣. The power rule and the rules for differentiating the sine and cosine are ddddsincosddcossinπ‘₯(π‘₯)=𝑛π‘₯,π‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=βˆ’π‘Žπ‘Žπ‘₯.

The given function 𝑦=(27π‘₯+27π‘₯)sincos is a composite, where 𝑒=π‘£οŠ¨ and 𝑣=27π‘₯+27π‘₯sincos. We begin by differentiating 𝑒 with respect to 𝑣 using the power rule and 𝑣 with respect to π‘₯ using the rules for differentiating the sine and cosine: ddddcossin𝑒𝑣=2𝑣,𝑣π‘₯=147π‘₯βˆ’147π‘₯.

Finally, substituting these expressions back into the chain rule where 𝑣=27π‘₯+27π‘₯sincos, we find the derivative of 𝑦 with respect to π‘₯ as ddsincoscossinsincoscossincossincossin𝑦π‘₯=(2(27π‘₯+27π‘₯))(147π‘₯βˆ’147π‘₯)=14(27π‘₯+27π‘₯)(27π‘₯βˆ’27π‘₯)=14ο€Ί47π‘₯βˆ’47π‘₯=56ο€Ί7π‘₯βˆ’7π‘₯.

We can simplify this further by applying the double angle trigonometric identity: coscossin2𝐴=π΄βˆ’π΄, where 𝐴=7π‘₯; we can simplify the derivative to obtain ddcos𝑦π‘₯=5614π‘₯.

We note that we could have also obtained this answer by first distributing the brackets in the given function, applying a trigonometric identity, and then differentiating the resulting expression.

Now, let’s consider an example where we will use the chain rule to find the derivative of a function composed of a square function and a tangent function.

Example 4: Finding the Derivative of a Trigonometric Function Using the Chain Rule

Find the derivative of the function 𝐽(πœƒ)=(π‘›πœƒ)tan.

Answer

In this example, we want to determine the derivative of a composite function using the chain rule and the general rule for differentiating the tangent functions.

Recall that the chain rule for a composite function 𝐽=𝑒(𝑣(πœƒ)) is given by ddddddπ½πœƒ=π‘’π‘£π‘£πœƒ, for differentiable functions 𝑒 and 𝑣. The power rule and the rules for differentiating the tangent function are ddddtansec𝑒(𝑒)=π‘Ÿπ‘’,πœƒ((π‘Žπœƒ))=π‘Ž(π‘Žπœƒ).

The given function 𝐽(πœƒ)=(π‘›πœƒ)tan is a composite function, where 𝑒=π‘£οŠ¨, where 𝑣=(π‘›πœƒ)tan. We begin by differentiating 𝑒 with respect to 𝑣 using the power rule and 𝑣 with respect to π‘₯ using the rules for differentiating the tangent function: ddddsec𝑒𝑣=2𝑣,π‘£πœƒ=𝑛(π‘›πœƒ).

Finally, substituting these expressions back into the chain rule where 𝑣=π‘›πœƒtan, we find the derivative of 𝐽 with respect to πœƒ as follows: ddtansecsectanπ½πœƒ=(2(π‘›πœƒ))𝑛(π‘›πœƒ)=2𝑛(π‘›πœƒ)(π‘›πœƒ).

We can also differentiate functions that contain trigonometric functions as products using the product rule.

Rule: The Product Rule

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

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

In the next example, we will find the derivative of a function that contains a product of a power and a sine function.

Example 5: Differentiating Functions Involving Trigonometric Ratios Using the Product Rule

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

Answer

In this example, we want to find the derivative of a product function that contains a trigonometric function using the product rule.

Recall that the product rule for the derivative of a product function 𝑦=𝑒(π‘₯)𝑣(π‘₯) is dddddd𝑦π‘₯=𝑒π‘₯𝑣(π‘₯)+𝑒(π‘₯)𝑣π‘₯, for differentiable functions 𝑒 and 𝑣. The power rule and the rules for differentiating the sine are ddddsincosπ‘₯(π‘₯)=𝑛π‘₯,π‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯).

The given function 𝑦=π‘₯(5π‘₯)sin is a product of two functions, where 𝑦=𝑒𝑣, where 𝑒=π‘₯ and 𝑣=(5π‘₯)sin. We begin by differentiating 𝑒 and 𝑣 with respect to π‘₯ using the power rule and the rule for differentiating the sine as ddddcos𝑒π‘₯=5π‘₯,𝑣π‘₯=5(5π‘₯).οŠͺ

Finally, substituting these expressions back into the product rule, we find the derivative of 𝑦 with respect to π‘₯ as ddsincoscossin𝑦π‘₯=ο€Ή5π‘₯((5π‘₯))+ο€Ήπ‘₯(5(5π‘₯))=5π‘₯(5π‘₯)+5π‘₯(5π‘₯).οŠͺοŠͺ

We can also find the derivative of trigonometric functions evaluated at a particular point, which is the same as the gradient of the tangent line to the curve at that point.

Now, let’s consider an example where we find the derivative of a function containing the product of a sine and tangent function using the product rule and evaluate the derivative at the given point.

Example 6: Differentiating Trigonometric Functions Using the Product Rule

Given 𝑦=(4π‘₯)(4π‘₯)sintan, find dd𝑦π‘₯ at π‘₯=πœ‹6.

Answer

In this example, we want to find the derivative of a product function that contains a trigonometric function using the product rule and then evaluate this at the given point.

Recall that the product rule for the derivative of a product function 𝑦=𝑒(π‘₯)𝑣(π‘₯) is dddddd𝑦π‘₯=𝑒π‘₯𝑣(π‘₯)+𝑒(π‘₯)𝑣π‘₯, for differentiable functions 𝑒 and 𝑣. The rules for differentiating the sine and tangent functions are ddsincosddtansecπ‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯),π‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯).

The given function 𝑦=(4π‘₯)(4π‘₯)sintan is a product of two functions, where 𝑦=𝑒𝑣, where 𝑒=(4π‘₯)sin and 𝑣=(4π‘₯)tan. We begin by differentiating 𝑒 and 𝑣 with respect to π‘₯ using the product rule and the rule for differentiating the sine and tangent as follows: ddcosddsec𝑒π‘₯=4(4π‘₯),𝑣π‘₯=4(4π‘₯).

Substituting these expressions back into the product rule, we find the derivative of 𝑦 with respect to π‘₯ as follows: ddcostansinseccossincossincossinsincoscossintansec𝑦π‘₯=(4(4π‘₯))((4π‘₯))+((4π‘₯))ο€Ή4(4π‘₯)=4(4π‘₯)Γ—(4π‘₯)(4π‘₯)+(4π‘₯)Γ—4(4π‘₯)=4(4π‘₯)+(4π‘₯)(4π‘₯)Γ—4(4π‘₯)=4(4π‘₯)+4(4π‘₯)(4π‘₯).

Finally, substituting π‘₯=πœ‹6, we find the derivative at this point: ddsintansecsintansec𝑦π‘₯|||=4ο€Ό4πœ‹6+4ο€Ό4πœ‹6οˆο€Ό4πœ‹6=4ο€Ό2πœ‹3+4ο€Ό2πœ‹3οˆο€Ό2πœ‹3=4Γ—βˆš32+4Γ—ο€»βˆ’βˆš3×(βˆ’2)=2√3+8√3=10√3.ο—οŠ²ο‘½οŽ₯

We can also differentiate functions that contain trigonometric functions as quotients, either in the numerator, denominator, or both, using the quotient rule, similar to how we found the derivative of the tangent function.

Now, let’s consider an example wherewe determine the derivative of a quotient of a linear function, in the numerator, and a tangent function, in the denominator.

Example 7: Finding the First Derivative of the Quotient of Trigonometric and Linear Functions Using the Quotient Rule

Differentiate 𝑦=4π‘₯5βˆ’π‘₯tan.

Answer

In this example, we want to find the derivative of a quotient function that contains a trigonometric function using the quotient rule.

Recall that the quotient rule for the derivative of a quotient function 𝑦=𝑒(π‘₯)𝑣(π‘₯) is dd𝑦π‘₯=𝑣(π‘₯)βˆ’π‘’(π‘₯)(𝑣(π‘₯)),ddddο‘ο—ο“ο—οŠ¨ for differentiable functions 𝑒 and 𝑣. The power rule and rules for differentiating the tangent function are ddddtansecπ‘₯(π‘₯)=𝑛π‘₯,π‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯).

The given function 𝑦=4π‘₯5βˆ’π‘₯tan is a quotient of two functions, where 𝑦=𝑒𝑣, where 𝑒=4π‘₯ and 𝑣=5βˆ’π‘₯tan. We begin by differentiating 𝑒 and 𝑣 with respect to π‘₯ using the power rule and the rule for differentiating the tangent function as follows: ddddsec𝑒π‘₯=4,𝑣π‘₯=βˆ’π‘₯.

Finally, substituting these expressions back into the quotient rule, we find the derivative of 𝑦 with respect to π‘₯: ddtansectantansectan𝑦π‘₯=(4)(5βˆ’π‘₯)βˆ’(4π‘₯)ο€Ήβˆ’π‘₯(5βˆ’π‘₯)=20βˆ’4π‘₯+4π‘₯π‘₯(5βˆ’π‘₯).

In the next example, we will find the derivative of a function containing the quotient of a cosine function in the numerator and the sine function in the denominator, using the quotient rule, and evaluate the derivative at the given point.

Example 8: Finding the First Derivative of the Quotient of Trigonometric Functions Using the Quotient Rule at a Point

Given that 𝑦=6π‘₯1βˆ’6π‘₯cossin, determine dd𝑦π‘₯ at π‘₯=πœ‹4.

Answer

In this example, we want to find the derivative of a quotient function that contains a trigonometric function using the quotient rule and then evaluate this at the given point.

Recall that the quotient rule for the derivative of a quotient function 𝑦=𝑒(π‘₯)𝑣(π‘₯) is dd𝑦π‘₯=𝑣(π‘₯)βˆ’π‘’(π‘₯)(𝑣(π‘₯)),ddddο‘ο—ο“ο—οŠ¨ for differentiable functions 𝑒 and 𝑣. The rules for differentiating the sine and cosine are ddsincosddcossinπ‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=βˆ’π‘Žπ‘Žπ‘₯.

The given function 𝑦=6π‘₯1βˆ’6π‘₯cossin is a quotient of two functions, where 𝑦=𝑒𝑣, where 𝑒=6π‘₯cos and 𝑣=1βˆ’6π‘₯sin. We begin by differentiating 𝑒 and 𝑣 with respect to π‘₯ using the power rule and the rule for differentiating the tangent function as follows: ddsinddcos𝑒π‘₯=βˆ’6(6π‘₯),𝑣π‘₯=βˆ’6(6π‘₯).

Substituting these expressions back into the quotient rule, we find the derivative of 𝑦 with respect to π‘₯: ddsinsincoscossinsinsincossin𝑦π‘₯=(βˆ’6(6π‘₯))(1βˆ’6π‘₯)βˆ’(6π‘₯)(βˆ’6(6π‘₯))(1βˆ’6π‘₯)=βˆ’6(6π‘₯)+6(6π‘₯)+6(6π‘₯)(1βˆ’6π‘₯).

Using the Pythagorean identity sincosοŠ¨οŠ¨π‘§+𝑧=1, we can simplify the expression as ddsinsinsinsinsin𝑦π‘₯=βˆ’6(6π‘₯)+6(1βˆ’6π‘₯)=6(1βˆ’(6π‘₯))(1βˆ’6π‘₯)=61βˆ’6π‘₯.

Finally, substituting π‘₯=πœ‹4, we find the derivative at this point as ddsinsin𝑦π‘₯|||=61βˆ’ο€»ο‡=61βˆ’ο€»ο‡=61βˆ’(βˆ’1)=62=3.ο—οŠ²οŠ¬οŽ„οŠͺοŠ©οŽ„οŠ¨ο‘½οŽ£

In the final example, we will determine the derivative of a function that contains a quotient with a product of a linear function and sine function in the numerator and another linear function in the denominator. We will make use of both the product rule and the quotient rule.

Example 9: Differentiating a Combination of Linear and Trigonometric Functions Using the Quotient Rule

Differentiate 𝑦=𝑑𝑑2+5𝑑sin.

Answer

In this example, we want to find the derivative of a quotient function that contains a trigonometric function and product function in the numerator, using the product and quotient rule.

Recall that the quotient rule for the derivative of a quotient function 𝑦=𝑒(𝑑)𝑣(𝑑) is 𝑦′(𝑑)=𝑒′(𝑑)𝑣(𝑑)βˆ’π‘’(𝑑)𝑣′(𝑑)(𝑣(𝑑)), for differentiable functions 𝑒 and 𝑣. Also, the product rule for the derivative of a product function 𝑒(𝑑)=𝑓(𝑑)𝑔(𝑑) is 𝑒′(𝑑)=𝑓′(𝑑)𝑔(𝑑)+𝑓(𝑑)𝑔′(𝑑), for differentiable functions 𝑓 and 𝑔. The power rule and the rules for differentiating the sine are ddddsincos𝑑(𝑑)=𝑛𝑑,𝑑((π‘Žπ‘‘))=π‘Ž(π‘Žπ‘‘).

The given function 𝑦=𝑑𝑑2+5𝑑sin is a quotient of two functions 𝑦=𝑒𝑣, where 𝑒(𝑑)=𝑑𝑑sin and 𝑣(𝑑)=2+5𝑑. We begin by differentiating 𝑒 and 𝑣 with respect to 𝑑.

We first note that the numerator 𝑒 is also a product of two functions 𝑒=𝑓𝑔, where 𝑓(𝑑)=𝑑 and 𝑔(𝑑)=𝑑sin and thus the derivative of 𝑒 can be found by applying the product rule with the derivatives of 𝑓 and 𝑔 using the power rule and the rule for differentiating the sine as follows: 𝑓′(𝑑)=1,𝑔′(𝑑)=𝑑,cos and thus the derivative of 𝑒 with respect to 𝑑 is 𝑒′(𝑑)=(1)(𝑑)+(𝑑)(𝑑)=𝑑+𝑑𝑑.sincossincos

The derivative of 𝑣 with respect to 𝑑 can be found by applying the power rule: 𝑣′(𝑑)=5.

Substituting these expressions back into the quotient rule, we find the derivative of 𝑦 with respect to 𝑑: 𝑦′(𝑑)=(𝑑+𝑑𝑑)(2+5𝑑)βˆ’(𝑑𝑑)(5)(2+5𝑑)=2𝑑+5𝑑𝑑+2𝑑𝑑+5π‘‘π‘‘βˆ’5𝑑𝑑(2+5𝑑)=5𝑑𝑑+2𝑑𝑑+2𝑑(2+5𝑑)=ο€Ή5𝑑+2𝑑𝑑+2𝑑(2+5𝑑).sincossinsinsincoscossincoscossincossin

Let’s now summarize a few key points from the explainer.

Key Points

  • We can determine the standard results for the derivatives of trigonometric functions from first principles using the definition of the derivative. It is sufficient to do this for the sine function and then find the results of the cosine and tangent function using trigonometric identities along with the chain and quotient rule. In particular, we find ddsincosddcossinddtansecπ‘₯(π‘₯)=π‘₯,π‘₯(π‘₯)=βˆ’π‘₯,π‘₯(π‘₯)=π‘₯.
  • The higher derivatives of sine and cosine have a cyclic pattern that repeats with period 4; the fourth derivative of either function gives you back the original function: ddοŠͺοŠͺ𝑓π‘₯=𝑓(π‘₯). This implies the following general rules for the π‘˜th derivative with π‘˜βˆˆβ„€: ddsinsinformodcosformodsinformodcosformodddcoscosformodsinformodcosformodsinformod()()()()π‘₯(π‘₯)=⎧βŽͺ⎨βŽͺ⎩π‘₯,π‘˜β‰‘0(4),π‘₯,π‘˜β‰‘1(4),βˆ’π‘₯,π‘˜β‰‘2(4),βˆ’π‘₯,π‘˜β‰‘3(4),π‘₯(π‘₯)=⎧βŽͺ⎨βŽͺ⎩π‘₯,π‘˜β‰‘0(4),βˆ’π‘₯,π‘˜β‰‘1(4),βˆ’π‘₯,π‘˜β‰‘2(4),π‘₯,π‘˜β‰‘3(4).
  • The general derivatives of the sine, cosine, and tangent functions can be found from the chain rule and are as follows: ddsincosddcossinddtansecπ‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯),π‘₯((π‘Žπ‘₯))=βˆ’π‘Ž(π‘Žπ‘₯),π‘₯((π‘Žπ‘₯))=π‘Ž(π‘Žπ‘₯), for π‘Žβˆˆβ„.
  • Using these standard results, we can find the derivatives of a large class of functions containing trigonometric functions using the linearity of the derivative, the chain rule, the product rule, and the quotient rule.

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