Explainer: Differentiation of Reciprocal Trigonometric Functions

In this explainer, we will learn how to find the derivatives of trigonometric functions, focusing on derivatives of cotangent, secant, and cosecant functions.

We will begin by recapping a few rules of differentiation. Firstly, we will recall the rules for differentiating the sine, cosine, and tangent functions.

Derivatives of Trigonometric Functions

The derivatives of the trigonometric functions are as follows: ddsincosddcossinddtansec𝑥(𝑎𝑥)=𝑎𝑎𝑥,𝑥(𝑎𝑥)=𝑎𝑎𝑥,𝑥(𝑎𝑥)=𝑎𝑎𝑥.

To differentiate the reciprocal trigonometric functions, we will appeal to the quotient rule of differentiation. Hence, before we begin, we will state both the quotient and the product rules which we will appeal to throughout the explainer.

Product Rule

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

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

Quotient Rule

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

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

Example 1: Derivative of the Secant Function

If 𝑦=22𝑥sec, determine the rate of change of 𝑦 when 𝑥=11𝜋6.

Answer

To find the rate of change of 𝑦 at 𝑥=11𝜋6, we will first find the derivative dd𝑦𝑥 and evaluate it at this point.

Using the definition of the secant function, seccos𝑥=1𝑥,

we can rewrite the expression for 𝑦 as 𝑦=22𝑥.cos

We can now apply the quotient rule, 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣, to 𝑦. Setting 𝑢=2 and 𝑣=2𝑥cos, we have 𝑢=0,𝑣=22𝑥.sin

Substituting these expressions into the quotient rule, we have ddcossincossincostansec𝑦𝑥=(2𝑥)(0)2(22𝑥)2𝑥=42𝑥2𝑥=42𝑥2𝑥.

Now that we have an expression for the rate of change of 𝑦 with respect to 𝑥, we can find its value when 𝑥=11𝜋6.

Hence, substituting in 𝑥=11𝜋6 into our expression gives ddtansec𝑦𝑥|||=4211𝜋6211𝜋6=43(2)=83.

Using the technique we used in the last example, it is possible to prove the general rule for the derivative of the secant function: ddsecsectan𝑥(𝑎𝑥)=𝑎𝑎𝑥𝑎𝑥.

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

Example 2: Derivative of the Cotangent Function

Determine dd𝑦𝑥 given that 𝑦=34𝑥+34𝑥coscot.

Answer

The derivative of a sum is the sum of the derivatives. Therefore, we can consider each term independently. Hence,

ddddcosddcot𝑦𝑥=𝑥(34𝑥)+𝑥(34𝑥).(1)

For the first of these terms, we can apply the standard results for differentiating trigonometric functions, ddcossin𝑥(𝑎𝑥)=𝑎𝑎𝑥, to get

ddcossinsin𝑥(34𝑥)=3(44𝑥)=124𝑥.(2)

As for the second term, we will use the definition of the cotangent function in terms of sine and cosine and apply the quotient rule. Recall cotcossin𝑥=𝑥𝑥; hence, we can write 34𝑥=34𝑥4𝑥.cotcossin

We can now apply the quotient rule, 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣, by setting 𝑢=34𝑥cos and 𝑣=4𝑥sin. We can now differentiate 𝑢 and 𝑣, using the standard results for the derivatives of sine and cosine to get 𝑢=124𝑥,𝑣=44𝑥.sincos

Substituting the expressions for 𝑢, 𝑣, 𝑢, and 𝑣 into the quotient rule gives ddcotsinsincoscossinsincossin𝑥(34𝑥)=(4𝑥)(124𝑥)(34𝑥)(44𝑥)4𝑥=124𝑥124𝑥4𝑥.

Using the Pythagorean identity for sine and cosine, we can simplify the numerator to get ddcotsincsc𝑥(34𝑥)=124𝑥=124𝑥.

Substituting this equation and equation (2) into equation (1), we get ddsincsc𝑦𝑥=124𝑥124𝑥.

In the last example, we showed that ddcotcsc𝑥(34𝑥)=124𝑥.

Using the same techniques, we can generalize the result to ddcotcsc𝑥(𝑎𝑥)=𝑎𝑎𝑥.

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

Example 3: Derivative of the Cosecant Function

Given that 𝑦=13(𝜋+5𝑥)csc, find dd𝑦𝑥.

Answer

When considering the derivative of a function like this, we could try to apply the trigonometric sum identity for the cosecant function. However, we can actually approach this in a different manner.

We will start by finding the derivative of csc5𝑥. Using the definition of the cosecant function, we have cscsin5𝑥=15𝑥.

To differentiate this, we can apply the quotient rule: 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣.

Setting 𝑢=1 and 𝑣=5𝑥sin, we have 𝑢=0,𝑣=55𝑥.cos

Therefore, ddcsccossincsccot𝑥(5𝑥)=055𝑥5𝑥=55𝑥5𝑥.

Now we know the derivative of csc5𝑥, we can consider the effect of adding 𝜋 to 5𝑥.

Consider the graph of a general function 𝑦=𝑓(𝑥). What is the effect of plotting 𝑦=𝑔(𝑥)=𝑓(𝑥𝑎) for some constant 𝑎? The graph of 𝑔 will be the graph of𝑓 translated by 𝑎 to the right as shown below.

Since we are interested in the derivative, we could ask a more interesting question: what is the effect of this on the derivative? Recall that the derivative at a point represents the slope of the curve at that point. Since translation does not deform the curve, we can conclude that the derivative of 𝑔 at a point 𝑥 will be equal to the derivative of 𝑓 at 𝑥𝑎. Hence, dd𝑥𝑓(𝑥𝑎)=𝑓(𝑥𝑎).

Therefore, given that the derivative of csc5𝑥 is 55𝑥5𝑥csccot, we can conclude that ddcsccsccot𝑥((𝜋+5𝑥))=5(𝜋+5𝑥)(𝜋+5𝑥).

Hence, ddcsccotcsccot𝑦𝑥=13(5(𝜋+5𝑥)(𝜋+5𝑥))=65(𝜋+5𝑥)(𝜋+5𝑥).

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

The methods used in the last three examples give us the following results for the derivatives of trigonometric functions.

Derivatives of Trigonometric Functions

The derivatives of the trigonometric functions are as follows: ddsincosddcossinddtansecddcotcscddsecsectanddcsccsccot𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏)(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏)(𝑎𝑥+𝑏).

Most of the time, when differentiating trigonometric functions, we can use these standard results without derivation. However, at times, you will be asked to derive these formulae. In all circumstances, it is extremely useful to commit these results to memory. This will dramatically improve your efficiency and increase your confidence in tackling differentiation questions.

In the last few examples, we will apply these standard results as we find the derivatives of more complex functions.

Example 4: Using the Product Rule with Reciprocal Trigonometric Functions

Given 𝑦=(𝑥+3)(9𝑥+𝑥)csc, find dd𝑦𝑥.

Answer

We can apply the product rule to the expression as is, or we could expand the parentheses before applying the product rule where necessary. Generally, it is best to find the simplest method. Sometimes by expanding parenthesis we arrive at a simpler expression. However, often it can be just as simple to use the product rule directly. In this case, we will use the product rule without expanding the parentheses. Recall the product rule: (𝑢𝑣)=𝑢𝑣+𝑢𝑣.

Setting 𝑢=(𝑥+3) and 𝑣=9𝑥+𝑥csc, we have 𝑢=1,𝑣=9𝑥𝑥.csccot

Substituting these expressions into the product rule, we have ddcsccsccotcsccotcsc𝑦𝑥=9𝑥+𝑥+(𝑥+3)(9𝑥𝑥)=18𝑥(𝑥+3)𝑥𝑥+𝑥+27.

Example 5: Differentiating Trigonometric Functions

If 𝑦=98𝑥8𝑥tansec, find dd𝑦𝑥.

Answer

To find this derivative, we will apply the product rule: (𝑢𝑣)=𝑢𝑣+𝑢𝑣.

Setting 𝑢=98𝑥tan and 𝑣=8𝑥sec, we have 𝑢=728𝑥,𝑣=88𝑥8𝑥.secsectan

Substituting these expressions into the product rule, we have ddtansectansecsectansecsec𝑦𝑥=(98𝑥)(88𝑥8𝑥)+(8𝑥)728𝑥=728𝑥8𝑥728𝑥.

The function in the previous example is actually the derivative of 988𝑥sec. Notice that the derivative is more complex than the original function. This is actually the case for all reciprocal trigonometric functions and the tangent function. This is in contrast to the derivatives of sine and cosine, which form a repeating pattern.

Example 6: Using the Quotient Rule with Reciprocal Trigonometric Functions

Given that 𝑦=(75𝑥+36𝑥)cotcsc, find dd𝑦𝑥.

Answer

For this example, we have a reciprocal function. Therefore, we can apply the quotient rule. There are alternative approaches. For example, we could use the chain rule. However, for the purpose of this example, we will use the quotient rule: 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣.

Setting 𝑢=1 and 𝑣=75𝑥+36𝑥cotcsc, we have 𝑢=0,𝑣=355𝑥186𝑥6𝑥.csccsccot

Substituting these expressions into the quotient rule, we have ddcsccsccotcotcsccsccotcsccotcsc𝑦𝑥=0355𝑥186𝑥6𝑥(75𝑥+36𝑥)=186𝑥6𝑥+355𝑥(75𝑥+36𝑥).

Key Points

  1. Using the quotient rule, we can find the derivatives of the reciprocal trigonometric functions.
  2. The derivatives of the trigonometric functions are as follows: ddsincosddcossinddtansecddcotcscddsecsectanddcsccsccot𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏)(𝑎𝑥+𝑏),𝑥((𝑎𝑥+𝑏))=𝑎(𝑎𝑥+𝑏)(𝑎𝑥+𝑏).
  3. Using these standard results in conjunction with the rules of differentiation, we can find the derivatives of a large class of functions.

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