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

Lesson Explainer: Differentiation of Trigonometric Functions Mathematics

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 sin𝑥, from first principles, and then use that result to also determine the derivatives of cos𝑥 and tan𝑥 using the chain and quotient rule.

Let’s first recall the definition of the derivative.

Definition: The Derivative

The derivative of a function 𝑓(𝑥) is defined as 𝑓(𝑥)=𝑓(𝑥+)𝑓(𝑥),lim at the points where the limit exists.

Substituting 𝑓(𝑥)=𝑥sin into the definition of the derivative, we have 𝑓(𝑥)=(𝑥+)𝑥.limsinsin

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

Using this identity, we can expand the sin(𝑥+) and rewrite the resulting expression using the sum and product rules of limits to obtain 𝑓(𝑥)=𝑥+𝑥𝑥=𝑥+𝑥(1)=𝑥×+𝑥×1=𝑥+𝑥1.limsincoscossinsinlimcossinsincoslimcossinsincoslimcoslimsinlimsinlimcos

Since both sin𝑥 and cos𝑥 are independent of , their limits will simply be sin𝑥 and cos𝑥 respectively. Hence, the derivative becomes 𝑓(𝑥)=𝑥+𝑥1.coslimsinsinlimcos

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

  1. limsin=1;
  2. limcos1=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 cos+1, which yields limcoslimcoscoscoslimcoscos1=1×+1+1=1(+1).

Using the Pythagorean identity for sine and cosine, cossin1=, we can rewrite the numerator in terms of sine and then apply the multiplicative law and take the resulting limits: limcoslimsincoslimsinsincoslimsinlimsincos1=(+1)=+1=+1=101+1=0.

Substituting the value of these two limits into 𝑓(𝑥), we obtain the derivative of 𝑓(𝑥)=𝑥sin as follows: 𝑓(𝑥)=𝑥+𝑥1=𝑥×1+𝑥×0=𝑥.coslimsinsinlimcoscossincos

We can also determine the derivative of 𝑓(𝑥)=𝑥cos from first principles in a similar way; however, it is better to determine the result from the cofunction identity cossin𝑥=𝜋2𝑥 and the chain rule with 𝑓(𝑥)=𝜋2𝑥sin.

Rule: The Chain Rule

Given two differentiable functions 𝑢(𝑥) and 𝑣(𝑥), the derivative of their composition 𝑢(𝑣(𝑥)) is given by dddddd𝑥(𝑢(𝑣(𝑥)))=𝑢𝑣𝑣𝑥.

This can be written succinctly using prime notation as follows: (𝑢(𝑣))=𝑢(𝑣)𝑣.

We can rewrite the function 𝑓(𝑥)=𝑥cos as 𝑓=𝑢sin with 𝑢=𝜋2𝑥, and the derivatives are ddcosdd𝑓𝑢=𝑢,𝑢𝑥=1.

Thus, using the chain rule, the derivative of 𝑓(𝑥)=𝑥cos is 𝑓(𝑥)=𝑓𝑢𝑢𝑥=𝜋2𝑥.ddddcos

We can rewrite the final result using the other cofunction identity, cossin𝜋2𝑥=𝑥, in reverse: 𝑓(𝑥)=𝑥.sin

Now, let’s consider higher derivatives of sin𝑥 and cos𝑥 that will form a cyclic pattern in the order of the derivatives. For 𝑓(𝑥)=𝑥sin, the first four derivatives are ddcosddsinddcosddsin𝑓𝑥=𝑥,𝑓𝑥=𝑥,𝑓𝑥=𝑥,𝑓𝑥=𝑥.

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 𝑘 can be written as ddsinformodcosformodsinformodcosformod()()𝑓𝑥=𝑥,𝑘0(4),𝑥,𝑘1(4),𝑥,𝑘2(4),𝑥,𝑘3(4).

Similarly, for 𝑓(𝑥)=𝑥cos, the first four derivatives are ddsinddcosddsinddcos𝑓𝑥=𝑥,𝑓𝑥=𝑥,𝑓𝑥=𝑥,𝑓𝑥=𝑥.

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 𝑘 can be written as ddcosformodsinformodcosformodsinformod()()𝑓𝑥=𝑥,𝑘0(4),𝑥,𝑘1(4),𝑥,𝑘2(4),𝑥,𝑘3(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(𝑥)=𝑥sin.

Answer

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

For 𝑓(𝑥)=𝑥sin, the first four derivatives are ddcosddsinddcosddsin𝑓𝑥=𝑥,𝑓𝑥=𝑥,𝑓𝑥=𝑥,𝑓𝑥=𝑥.

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 𝑘th derivative with 𝑘 is ddsinformodcosformodsinformodcosformod()()𝑓𝑥=𝑥,𝑘0(4),𝑥,𝑘1(4),𝑥,𝑘2(4),𝑥,𝑘3(4).

Since 33=4×8+1, we have 331(4)mod. Hence, the thirty-third derivative is equivalent to the first derivative of 𝑓(𝑥)=𝑥sin and we have ddddcos()()𝑓𝑥=𝑓𝑥=𝑥.

The derivative of 𝑓(𝑥)=𝑥tan can be found by using the quotient rule; by using the trigonometric identity that tansincos𝑥=𝑥𝑥, we can express 𝑓(𝑥) as 𝑓(𝑥)=𝑥𝑥.sincos

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

Rule: The Quotient Rule

Given two differentiable functions 𝑢(𝑥) and 𝑣(𝑥), the derivative of their quotient 𝑢(𝑥)𝑣(𝑥) is given by dd𝑥𝑢(𝑥)𝑣(𝑥)=𝑣(𝑥)𝑢(𝑥)(𝑣(𝑥)).dddd

This can be written succinctly using prime notation as follows: 𝑢𝑣=𝑢𝑣𝑢𝑣𝑣.

Setting 𝑢=𝑥sin and 𝑣=𝑥cos, we begin by differentiating 𝑢 and 𝑣 to find expressions for 𝑢 and 𝑣 from previously established results: 𝑢=𝑥,𝑣=𝑥.cossin

Substituting these expressions into the quotient rule, we have 𝑓(𝑥)=𝑥(𝑥)(𝑥)𝑥(𝑥)=𝑥+𝑥𝑥.coscossinsincoscossincos

Using the Pythagorean identity for sine and cosine, we can simplify the numerator. Hence, 𝑓(𝑥)=1𝑥=𝑥.cossec

In general, we can establish a rule for derivatives of the function 𝑓(𝑥)=(𝑎𝑥)sin using the chain rule; we can rewrite this as 𝑓=𝑢sin, where 𝑢=𝑎𝑥, and using ddcosdd𝑓𝑢=𝑢,𝑢𝑥=𝑎 the derivative becomes 𝑓(𝑥)=𝑓𝑢𝑢𝑥=𝑎(𝑎𝑥).ddddcos

Similarly, for 𝑓(𝑥)=(𝑎𝑥)cos, using the chain rule by rewriting the function as 𝑓=𝑢cos, where 𝑢=𝑎𝑥, we obtain the general derivative 𝑓(𝑥)=𝑎(𝑎𝑥).sin

The derivative of 𝑓(𝑥)=(𝑎𝑥)tan can also be found using the chain rule by rewriting the function as 𝑓=𝑢tan, where 𝑢=𝑎𝑥, or using the quotient rule with the derivatives of the general sine and cosine functions to obtain 𝑓(𝑥)=𝑎(𝑎𝑥).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: ddsincosddcossinddtansec𝑥((𝑎𝑥))=𝑎(𝑎𝑥),𝑥((𝑎𝑥))=𝑎(𝑎𝑥),𝑥((𝑎𝑥))=𝑎(𝑎𝑥), for 𝑎.

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: ddddddfor𝑥(𝑎𝑓+𝑏𝑔)=𝑎𝑓𝑥+𝑏𝑔𝑥,𝑎,𝑏.

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=2𝑥+6𝑥2+𝜋4sincos, find dd𝑦𝑥.

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 ddddsincos𝑥(𝑥)=𝑛𝑥,𝑥((𝑎𝑥))=𝑎(𝑎𝑥).

Applying these rules, we can calculate the first derivative of 𝑦=2𝑥+6𝑥2+𝜋4sincos as ddddddsinddcoscoscos𝑦𝑥=2𝑥𝑥+6𝑥𝑥2+𝑥𝜋4=2×4𝑥+6×12𝑥2+0=8𝑥+3𝑥2.

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=(27𝑥+27𝑥)sincos, find dd𝑦𝑥.

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=𝑢(𝑣(𝑥)) is given by dddddd𝑦𝑥=𝑢𝑣𝑣𝑥, for differentiable functions 𝑢 and 𝑣. The power rule and the rules for differentiating the sine and cosine are ddddsincosddcossin𝑥(𝑥)=𝑛𝑥,𝑥(𝑎𝑥)=𝑎𝑎𝑥,𝑥(𝑎𝑥)=𝑎𝑎𝑥.

The given function 𝑦=(27𝑥+27𝑥)sincos is a composite, where 𝑢=𝑣 and 𝑣=27𝑥+27𝑥sincos. We begin by differentiating 𝑢 with respect to 𝑣 using the power rule and 𝑣 with respect to 𝑥 using the rules for differentiating the sine and cosine: ddddcossin𝑢𝑣=2𝑣,𝑣𝑥=147𝑥147𝑥.

Finally, substituting these expressions back into the chain rule where 𝑣=27𝑥+27𝑥sincos, we find the derivative of 𝑦 with respect to 𝑥 as ddsincoscossinsincoscossincossincossin𝑦𝑥=(2(27𝑥+27𝑥))(147𝑥147𝑥)=14(27𝑥+27𝑥)(27𝑥27𝑥)=1447𝑥47𝑥=567𝑥7𝑥.

We can simplify this further by applying the double angle trigonometric identity: coscossin2𝐴=𝐴𝐴, where 𝐴=7𝑥; we can simplify the derivative to obtain ddcos𝑦𝑥=5614𝑥.

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(𝜃)=(𝑛𝜃)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=𝑢(𝑣(𝜃)) is given by dddddd𝐽𝜃=𝑢𝑣𝑣𝜃, for differentiable functions 𝑢 and 𝑣. The power rule and the rules for differentiating the tangent function are ddddtansec𝑢(𝑢)=𝑟𝑢,𝜃((𝑎𝜃))=𝑎(𝑎𝜃).

The given function 𝐽(𝜃)=(𝑛𝜃)tan is a composite function, where 𝑢=𝑣, where 𝑣=(𝑛𝜃)tan. We begin by differentiating 𝑢 with respect to 𝑣 using the power rule and 𝑣 with respect to 𝑥 using the rules for differentiating the tangent function: ddddsec𝑢𝑣=2𝑣,𝑣𝜃=𝑛(𝑛𝜃).

Finally, substituting these expressions back into the chain rule where 𝑣=𝑛𝜃tan, we find the derivative of 𝐽 with respect to 𝜃 as follows: ddtansecsectan𝐽𝜃=(2(𝑛𝜃))𝑛(𝑛𝜃)=2𝑛(𝑛𝜃)(𝑛𝜃).

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

Rule: The Product Rule

Given two differentiable functions 𝑢(𝑥) and 𝑣(𝑥), the derivative of their product 𝑢(𝑥)𝑣(𝑥) is given by dddddd𝑥(𝑢(𝑥)𝑣(𝑥))=𝑢𝑥𝑣(𝑥)+𝑢(𝑥)𝑣𝑥.

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

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=𝑥(5𝑥)sin, determine dd𝑦𝑥.

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=𝑢(𝑥)𝑣(𝑥) is dddddd𝑦𝑥=𝑢𝑥𝑣(𝑥)+𝑢(𝑥)𝑣𝑥, for differentiable functions 𝑢 and 𝑣. The power rule and the rules for differentiating the sine are ddddsincos𝑥(𝑥)=𝑛𝑥,𝑥((𝑎𝑥))=𝑎(𝑎𝑥).

The given function 𝑦=𝑥(5𝑥)sin is a product of two functions, where 𝑦=𝑢𝑣, where 𝑢=𝑥 and 𝑣=(5𝑥)sin. We begin by differentiating 𝑢 and 𝑣 with respect to 𝑥 using the power rule and the rule for differentiating the sine as ddddcos𝑢𝑥=5𝑥,𝑣𝑥=5(5𝑥).

Finally, substituting these expressions back into the product rule, we find the derivative of 𝑦 with respect to 𝑥 as ddsincoscossin𝑦𝑥=5𝑥((5𝑥))+𝑥(5(5𝑥))=5𝑥(5𝑥)+5𝑥(5𝑥).

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=(4𝑥)(4𝑥)sintan, find dd𝑦𝑥 at 𝑥=𝜋6.

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=𝑢(𝑥)𝑣(𝑥) is dddddd𝑦𝑥=𝑢𝑥𝑣(𝑥)+𝑢(𝑥)𝑣𝑥, for differentiable functions 𝑢 and 𝑣. The rules for differentiating the sine and tangent functions are ddsincosddtansec𝑥((𝑎𝑥))=𝑎(𝑎𝑥),𝑥((𝑎𝑥))=𝑎(𝑎𝑥).

The given function 𝑦=(4𝑥)(4𝑥)sintan is a product of two functions, where 𝑦=𝑢𝑣, where 𝑢=(4𝑥)sin and 𝑣=(4𝑥)tan. We begin by differentiating 𝑢 and 𝑣 with respect to 𝑥 using the product rule and the rule for differentiating the sine and tangent as follows: ddcosddsec𝑢𝑥=4(4𝑥),𝑣𝑥=4(4𝑥).

Substituting these expressions back into the product rule, we find the derivative of 𝑦 with respect to 𝑥 as follows: ddcostansinseccossincossincossinsincoscossintansec𝑦𝑥=(4(4𝑥))((4𝑥))+((4𝑥))4(4𝑥)=4(4𝑥)×(4𝑥)(4𝑥)+(4𝑥)×4(4𝑥)=4(4𝑥)+(4𝑥)(4𝑥)×4(4𝑥)=4(4𝑥)+4(4𝑥)(4𝑥).

Finally, substituting 𝑥=𝜋6, we find the derivative at this point: ddsintansecsintansec𝑦𝑥|||=44𝜋6+44𝜋64𝜋6=42𝜋3+42𝜋32𝜋3=4×32+4×3×(2)=23+83=103.

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=4𝑥5𝑥tan.

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=𝑢(𝑥)𝑣(𝑥) is dd𝑦𝑥=𝑣(𝑥)𝑢(𝑥)(𝑣(𝑥)),dddd for differentiable functions 𝑢 and 𝑣. The power rule and rules for differentiating the tangent function are ddddtansec𝑥(𝑥)=𝑛𝑥,𝑥((𝑎𝑥))=𝑎(𝑎𝑥).

The given function 𝑦=4𝑥5𝑥tan is a quotient of two functions, where 𝑦=𝑢𝑣, where 𝑢=4𝑥 and 𝑣=5𝑥tan. We begin by differentiating 𝑢 and 𝑣 with respect to 𝑥 using the power rule and the rule for differentiating the tangent function as follows: ddddsec𝑢𝑥=4,𝑣𝑥=𝑥.

Finally, substituting these expressions back into the quotient rule, we find the derivative of 𝑦 with respect to 𝑥: ddtansectantansectan𝑦𝑥=(4)(5𝑥)(4𝑥)𝑥(5𝑥)=204𝑥+4𝑥𝑥(5𝑥).

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=6𝑥16𝑥cossin, determine dd𝑦𝑥 at 𝑥=𝜋4.

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=𝑢(𝑥)𝑣(𝑥) is dd𝑦𝑥=𝑣(𝑥)𝑢(𝑥)(𝑣(𝑥)),dddd for differentiable functions 𝑢 and 𝑣. The rules for differentiating the sine and cosine are ddsincosddcossin𝑥(𝑎𝑥)=𝑎𝑎𝑥,𝑥(𝑎𝑥)=𝑎𝑎𝑥.

The given function 𝑦=6𝑥16𝑥cossin is a quotient of two functions, where 𝑦=𝑢𝑣, where 𝑢=6𝑥cos and 𝑣=16𝑥sin. We begin by differentiating 𝑢 and 𝑣 with respect to 𝑥 using the power rule and the rule for differentiating the tangent function as follows: ddsinddcos𝑢𝑥=6(6𝑥),𝑣𝑥=6(6𝑥).

Substituting these expressions back into the quotient rule, we find the derivative of 𝑦 with respect to 𝑥: ddsinsincoscossinsinsincossin𝑦𝑥=(6(6𝑥))(16𝑥)(6𝑥)(6(6𝑥))(16𝑥)=6(6𝑥)+6(6𝑥)+6(6𝑥)(16𝑥).

Using the Pythagorean identity sincos𝑧+𝑧=1, we can simplify the expression as ddsinsinsinsinsin𝑦𝑥=6(6𝑥)+6(16𝑥)=6(1(6𝑥))(16𝑥)=616𝑥.

Finally, substituting 𝑥=𝜋4, we find the derivative at this point as ddsinsin𝑦𝑥|||=61=61=61(1)=62=3.

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=𝑡𝑡2+5𝑡sin.

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=𝑢(𝑡)𝑣(𝑡) is 𝑦(𝑡)=𝑢(𝑡)𝑣(𝑡)𝑢(𝑡)𝑣(𝑡)(𝑣(𝑡)), for differentiable functions 𝑢 and 𝑣. Also, the product rule for the derivative of a product function 𝑢(𝑡)=𝑓(𝑡)𝑔(𝑡) is 𝑢(𝑡)=𝑓(𝑡)𝑔(𝑡)+𝑓(𝑡)𝑔(𝑡), for differentiable functions 𝑓 and 𝑔. The power rule and the rules for differentiating the sine are ddddsincos𝑡(𝑡)=𝑛𝑡,𝑡((𝑎𝑡))=𝑎(𝑎𝑡).

The given function 𝑦=𝑡𝑡2+5𝑡sin is a quotient of two functions 𝑦=𝑢𝑣, where 𝑢(𝑡)=𝑡𝑡sin and 𝑣(𝑡)=2+5𝑡. We begin by differentiating 𝑢 and 𝑣 with respect to 𝑡.

We first note that the numerator 𝑢 is also a product of two functions 𝑢=𝑓𝑔, where 𝑓(𝑡)=𝑡 and 𝑔(𝑡)=𝑡sin and thus the derivative of 𝑢 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(𝑡)=1,𝑔(𝑡)=𝑡,cos and thus the derivative of 𝑢 with respect to 𝑡 is 𝑢(𝑡)=(1)(𝑡)+(𝑡)(𝑡)=𝑡+𝑡𝑡.sincossincos

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

Substituting these expressions back into the quotient rule, we find the derivative of 𝑦 with respect to 𝑡: 𝑦(𝑡)=(𝑡+𝑡𝑡)(2+5𝑡)(𝑡𝑡)(5)(2+5𝑡)=2𝑡+5𝑡𝑡+2𝑡𝑡+5𝑡𝑡5𝑡𝑡(2+5𝑡)=5𝑡𝑡+2𝑡𝑡+2𝑡(2+5𝑡)=5𝑡+2𝑡𝑡+2𝑡(2+5𝑡).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 ddsincosddcossinddtansec𝑥(𝑥)=𝑥,𝑥(𝑥)=𝑥,𝑥(𝑥)=𝑥.
  • 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: dd𝑓𝑥=𝑓(𝑥). This implies the following general rules for the 𝑘th derivative with 𝑘: ddsinsinformodcosformodsinformodcosformodddcoscosformodsinformodcosformodsinformod()()()()𝑥(𝑥)=𝑥,𝑘0(4),𝑥,𝑘1(4),𝑥,𝑘2(4),𝑥,𝑘3(4),𝑥(𝑥)=𝑥,𝑘0(4),𝑥,𝑘1(4),𝑥,𝑘2(4),𝑥,𝑘3(4).
  • The general derivatives of the sine, cosine, and tangent functions can be found from the chain rule and are as follows: ddsincosddcossinddtansec𝑥((𝑎𝑥))=𝑎(𝑎𝑥),𝑥((𝑎𝑥))=𝑎(𝑎𝑥),𝑥((𝑎𝑥))=𝑎(𝑎𝑥), for 𝑎.
  • 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.

Download the Nagwa Classes App

Attend sessions, chat with your teacher and class, and access class-specific questions. Download Nagwa Classes app today!

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.