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

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

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

These functions are defined as the reciprocal of the standard trigonometric functions: sine, cosine, and tangent, and hence they are called the reciprocal trigonometric functions. Let us recall the definition of these functions.

Definition: Reciprocal Trigonometric Functions

The reciprocal trigonometric functions are as follows: seccosifforanycscsinifforanycotcossinifforany𝑥=1𝑥𝑥𝜋2+𝑛𝜋𝑛,𝑥=1𝑥𝑥𝑛𝜋𝑛,𝑥=𝑥𝑥𝑥𝑛𝜋𝑛.

Note that cottan𝑥=1𝑥 if 𝑥𝜋2+𝑛𝜋 for any 𝑛.

To compute the derivatives of these reciprocal trigonometric functions, we can begin with the derivatives of the standard trigonometric functions and apply the quotient rule to the reciprocal expression. Although we can always obtain the derivatives of reciprocal trigonometric functions this way, it is useful to know the formulas for these derivatives so that we do not need to derive these expressions each time. We recall the derivatives for the sine, cosine, and tangent functions.

Rule: Derivatives of Trigonometric Functions

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

Let us begin with the derivative of the secant function. From the definition of the secant function, we can write

ddsecddcos𝑥𝑥=𝑥1𝑥.(1)

We will apply the quotient rule to this expression to find the derivative.

Rule: Quotient Rule

Given differentiable functions 𝑓(𝑥) and 𝑔(𝑥), ddif𝑥𝑓(𝑥)𝑔(𝑥)=𝑓(𝑥)𝑔(𝑥)𝑓(𝑥)𝑔(𝑥)(𝑔(𝑥))𝑔(𝑥)0.

Applying the quotient rule to the right-hand side of equation (1), we have ddcoscoscoscos𝑥1𝑥=(1)𝑥1(𝑥)(𝑥).

By the constant rule, we know that (1)=0. We also know that (𝑥)=𝑥cossin. Hence, the right-hand side of the equation above simplifies to 0𝑥1(𝑥)(𝑥)=𝑥𝑥.cossincossincos

We can further simplify this expression using the definitions tansincos𝑥=𝑥𝑥 and seccos𝑥=1𝑥. The right-hand side of the equation above can be written as sincoscostansec𝑥𝑥×1𝑥=𝑥𝑥.

This leads to the derivative of the secant function.

Rule: Derivative of the Secant Function

If 𝑥𝜋2+𝑛𝜋 for any 𝑛, ddsecsectan𝑥𝑥=𝑥𝑥.

Below, we can observe the graph of the secant function plotted together with the graph of its derivative.

We can combine this derivative rule with the chain rule. Let us recall the chain rule.

Rule: Chain Rule

Given two differentiable functions 𝑓(𝑢) and 𝑔(𝑥), dd𝑥(𝑓(𝑔(𝑥)))=𝑓(𝑔(𝑥))𝑔(𝑥).

In our first example, we will apply the chain rule with the rule for differentiation of the secant function to find the derivative of a given function at a point.

Example 1: Differentiating Trigonometric Functions

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

Answer

Recall that the rate of change of 𝑦 is given by the derivative dd𝑦𝑥. In this example, we can find the rate of change of 𝑦 when 𝑥=11𝜋6 by first finding the derivative dd𝑦𝑥 and then evaluating the derivative at 𝑥=11𝜋6.

The function that we want to differentiate involves the secant function. Hence, we begin by recalling the derivative of the secant function: ddsecsectan𝑥𝑥=𝑥𝑥.

To find the derivative dd𝑦𝑥, we need to differentiate the function 22𝑥sec. The constant 2 can be factored outside of the derivative by the constant multiple rule, which means ddddsecddsec𝑦𝑥=𝑥(22𝑥)=2𝑥(2𝑥).

We will apply the chain rule to take care of the expression 2𝑥 inside the secant function. We recall the chain rule: given two differentiable functions 𝑓(𝑢) and 𝑔(𝑥), dd𝑥𝑓(𝑔(𝑥))=𝑓(𝑔(𝑥))𝑔(𝑥).

We can see that the outside function of sec2𝑥 is sec𝑢, and the inside function is 2𝑥. Hence, 𝑓(𝑢)=𝑢sec and 𝑔(𝑥)=2𝑥. We know that 𝑓(𝑢)=𝑢𝑢=𝑢𝑢.ddsecsectan

Applying the power rule to 𝑔(𝑥), we obtain 𝑔(𝑥)=𝑥2𝑥=2.dd

Hence, applying the chain rule to sec2𝑥 leads to 𝑓(𝑔(𝑥))𝑔(𝑥)=2𝑥2𝑥×2=22𝑥2𝑥.sectansectan

Finally, remembering the constant 2 in front of this derivative, we obtain ddsectansectan𝑦𝑥=2(22𝑥2𝑥)=42𝑥2𝑥.

Now, we need to evaluate this expression at 𝑥=11𝜋6. Substituting this value into dd𝑦𝑥, we obtain ddsectan𝑦𝑥|||=411𝜋311𝜋3.

We note that 11𝜋3>2𝜋; hence, we can find an equivalent angle by subtracting 2𝜋 radians from this angle: 11𝜋32𝜋=11𝜋36𝜋3=5𝜋3.

This means that ddsectan𝑦𝑥|||=45𝜋35𝜋3.

Now, we see that 5𝜋3 is a special angle in the unit circle that has the following trigonometric ratios: sincos5𝜋3=32,5𝜋3=12.

Since secsin𝜃=1𝜃 and tansincos𝜃=𝜃𝜃 for any angle 𝜃, we have sectan5𝜋3=1=2,5𝜋3==3.

Finally, substituting these values above, we obtain dd𝑦𝑥|||=4×2×3=83.

Hence, the rate of change of 𝑦 at the given 𝑥 value is 83.

In the last example, we combined this derivative rule for the secant function with the chain. We can also combine this rule with the product rule.

Rule: Product Rule

Given two differentiable functions 𝑓(𝑥) and 𝑔(𝑥), dd𝑥(𝑓(𝑥)𝑔(𝑥))=𝑓(𝑥)𝑔(𝑥)+𝑓(𝑥)𝑔(𝑥).

In the next example, we will use the derivative rule for the secant function together with the product rule and the chain rule.

Example 2: Differentiating Trigonometric Functions Involving Trigonometric Ratios Using the Product Rule

If 𝑦=98𝑥8𝑥tansec, find dd𝑦𝑥.

Answer

The function that we want to differentiate involves the tangent and secant functions, so we can begin by recalling these derivatives: ddtansecddsecsectan𝑥𝑥=𝑥,𝑥𝑥=𝑥𝑥.

To find the derivative dd𝑦𝑥, we need to differentiate the function 98𝑥8𝑥tansec that is a product of two functions. The constant 9 can be factored outside of the derivative by the constant multiple rule, which leads to ddddtansecddtansec𝑦𝑥=𝑥(98𝑥8𝑥)=9𝑥(8𝑥8𝑥).

Now, to differentiate tansec8𝑥8𝑥, we notice that this is a product of two functions. Hence, we recall the product rule: given two differentiable functions 𝑓(𝑥) and 𝑔(𝑥), dd𝑥(𝑓(𝑥)𝑔(𝑥))=𝑓(𝑥)𝑔(𝑥)+𝑓(𝑥)𝑔(𝑥).

Applying the product rule, we can write ddtansectansectansec𝑥(8𝑥8𝑥)=(8𝑥)8𝑥+8𝑥(8𝑥).

Finally, we need to compute the derivatives (8𝑥)tan and (8𝑥)sec; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions 𝑓(𝑢) and 𝑔(𝑥), dd𝑥𝑓(𝑔(𝑥))=𝑓(𝑔(𝑥))𝑔(𝑥).

For tan8𝑥, the outside function is 𝑓(𝑢)=𝑢tan, while the inside function is 𝑔(𝑥)=8𝑥. Since we know 𝑓(𝑢)=𝑢sec and 𝑔(𝑥)=8 by the power rule, we can write the following using the chain rule: (8𝑥)=8𝑥×8=88𝑥.tansecsec

For sec8𝑥, the outside function is 𝑓(𝑢)=𝑢sec, while the inside function is 𝑔(𝑥)=8𝑥. Using 𝑓(𝑢)=𝑢𝑢sectan and 𝑔(𝑥)=8, (8𝑥)=8𝑥8𝑥×8=88𝑥8𝑥.secsectansectan

Substituting these expressions into the product rule above, ddtansecsecsectansectansecsectan𝑥(8𝑥8𝑥)=88𝑥8𝑥+8𝑥(88𝑥8𝑥)=88𝑥+88𝑥8𝑥.

We must not forget the constant 9 that we had factored outside of the derivative. Hence, we obtain ddsecsectansecsectansectansec𝑦𝑥=988𝑥+88𝑥8𝑥=728𝑥728𝑥8𝑥=728𝑥8𝑥728𝑥.

In previous examples, we used the derivative rule for the secant function together with the chain rule and product rule to differentiate given functions. Let us now find the formula for the derivative of the cosecant function. From the definition of the cosecant function, we can write ddcscddsin𝑥𝑥=𝑥1𝑥.

Applying the quotient rule, ddsinsinsinsin𝑥1𝑥=(1)𝑥1(𝑥)(𝑥).

By the constant rule, we know that (1)=0. We also know that (𝑥)=𝑥sincos. Hence, the right-hand side of the equation above simplifies to 0𝑥1(𝑥)(𝑥)=𝑥𝑥.sincossincossin

We can further simplify this expression using the definitions cotcossin𝑥=𝑥𝑥 and cscsin𝑥=1𝑥. The right-hand side of the equation above can be written as 𝑥𝑥×1𝑥=𝑥𝑥.cossinsincotcsc

This leads to the derivative of the cosecant function.

Rule: Derivative of the Cosecant Function

If 𝑥𝑛𝜋 for any 𝑛, ddcsccsccot𝑥𝑥=𝑥𝑥.

Below, we can observe the graph of the cosecant function plotted together with the graph of its derivative.

Let us consider an example where we will apply this rule together with the product rule.

Example 3: Differentiating Functions Involving Reciprocal Trigonometric Ratios Using the Product Rule

Given 𝑦=(𝑥+3)(9𝑥+𝑥)csc, find dd𝑦𝑥.

Answer

The function that we want to differentiate involves the cosecant function, so we can begin by recalling the derivative of the cosecant function: ddcsccsccot𝑥𝑥=𝑥𝑥.

To find the derivative dd𝑦𝑥, we need to differentiate the function (𝑥+3)(9𝑥+𝑥)csc that is a product of two functions. We recall the product rule: given two differentiable functions 𝑓(𝑥) and 𝑔(𝑥), dd𝑥(𝑓(𝑥)𝑔(𝑥))=𝑓(𝑥)𝑔(𝑥)+𝑓(𝑥)𝑔(𝑥).

Applying the product rule, we can write ddddcsccsccsc𝑦𝑥=𝑥((𝑥+3)(9𝑥+𝑥))=(𝑥+3)(9𝑥+𝑥)+(𝑥+3)(9𝑥+𝑥).

The first factor (𝑥+3) is the derivative of a polynomial, so we can compute this derivative by the power rule: (𝑥+3)=1+0=1.

The last factor (9𝑥+𝑥)csc is the derivative of a function that is the sum of the cosecant function and a polynomial. Then, (9𝑥+𝑥)=(9𝑥)+(𝑥)=9𝑥𝑥.csccsccsccot

Substituting these expressions into the product rule, we have ddcsccsccotcsccsccotcsccotcsc𝑦𝑥=1(9𝑥+𝑥)+(𝑥+3)(9𝑥𝑥)=9𝑥+𝑥+9𝑥+27(𝑥+3)𝑥𝑥=18𝑥(𝑥+3)𝑥𝑥+𝑥+27.

Let us consider another example where we will use the derivative formula for the cosecant function together with the chain rule.

Example 4: Differentiating Reciprocal Trigonometric Functions Using the Chain Rule

Given that 𝑦=13(𝜋+5𝑥)csc, find dd𝑦𝑥.

Answer

The function that we want to differentiate involves the cosecant function. Hence, we begin by recalling the derivative of the cosecant function: ddcsccsccot𝑥𝑥=𝑥𝑥.

To find the derivative dd𝑦𝑥, we need to differentiate the function 13(𝜋+5𝑥)csc. The constant 13 can be factored outside of the derivative by the constant multiple rule, which means ddddcscddcsc𝑦𝑥=𝑥(13(𝜋+5𝑥))=13𝑥(𝜋+5𝑥).

Since csc(𝜋+5𝑥) is a composition of two functions, we recall the chain rule: given two differentiable functions 𝑓(𝑢) and 𝑔(𝑥), dd𝑥𝑓(𝑔(𝑥))=𝑓(𝑔(𝑥))𝑔(𝑥).

We can see that the outside function of csc(𝜋+5𝑥) is 𝑓(𝑢)=𝑢csc, and the inside function is 𝑔(𝑥)=𝜋+5𝑥. Hence, 𝑓(𝑢)=𝑢𝑢csccot. For the inside function 𝑔, we note that (𝜋)=0 by the constant rule, and (5𝑥)=5 by the power rule. This leads to 𝑔(𝑥)=5. Substituting these expressions into the chain rule, we obtain ddcsccsccotcsccot𝑥(𝜋+5𝑥)=(𝜋+5𝑥)(𝜋+5𝑥)×5=5(𝜋+5𝑥)(𝜋+5𝑥).

Finally, remembering the constant 13 in front of this derivative, we obtain ddcsccotcsccot𝑦𝑥=13(5(𝜋+5𝑥)(𝜋+5𝑥))=65(𝜋+5𝑥)(𝜋+5𝑥).

So far, we have discussed various derivative problems involving the secant and cosecant functions. Let us now turn our attention to the last remaining reciprocal trigonometric function, the cotangent. From the definition of the cotangent function, we can write ddcotddcossin𝑥𝑥=𝑥𝑥𝑥.

Applying the quotient rule, ddcossincossincossinsin𝑥𝑥𝑥=(𝑥)𝑥𝑥(𝑥)(𝑥).

We know that (𝑥)=𝑥,(𝑥)=𝑥.cossinsincos

Hence, the derivative of cot𝑥 can be written as 𝑥×𝑥𝑥×𝑥(𝑥)=𝑥+𝑥𝑥.sinsincoscossinsincossin

We can apply the trigonometric identity sincos𝑥+𝑥=1 to write this derivative as 1𝑥sin. Finally, using the definition cscsin𝑥=1𝑥, the resulting expression can be written as 𝑥csc. This leads to the derivative of the cotangent function.

Rule: Derivative of the Cotangent Function

If 𝑥𝑛𝜋 for any 𝑛, ddcotcsc𝑥𝑥=𝑥.

Below, we can observe the graph of the cosecant function plotted together with the graph of its derivative.

Let us consider an example where we need to apply this rule to find a derivative.

Example 5: Differentiating a Combination of Trigonometric Functions

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

Answer

The function that we want to differentiate involves the cosine and cotangent functions, so we can begin by recalling these derivatives: ddcossinddcotcsc𝑥𝑥=𝑥,𝑥𝑥=𝑥.

To find dd𝑦𝑥, we need to differentiate the function 34𝑥+34𝑥coscot. The sum can be split up using the sum/difference rule, and the constants 3 and 3 can be factored outside of each derivative leading to ddddcoscotddcosddcotddcosddcot𝑦𝑥=𝑥(34𝑥+34𝑥)=𝑥(34𝑥)+𝑥(34𝑥)=3𝑥(4𝑥)+3𝑥(4𝑥).

Now we need to compute the derivatives (4𝑥)cos and (4𝑥)cot; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions 𝑓(𝑢) and 𝑔(𝑥), dd𝑥𝑓(𝑔(𝑥))=𝑓(𝑔(𝑥))𝑔(𝑥).

For cos4𝑥, the outside function is 𝑓(𝑢)=𝑢cos, while the inside function is 𝑔(𝑥)=4𝑥. Since we know 𝑓(𝑢)=𝑢sin and 𝑔(𝑥)=4 by the power rule, we can write the following using the chain rule: (4𝑥)=4𝑥×4=44𝑥.cossinsin

For cot4𝑥, the outside function is 𝑓(𝑢)=𝑢cot, while the inside function is 𝑔(𝑥)=4𝑥. Using 𝑓(𝑢)=𝑢csc and 𝑔(𝑥)=4, (4𝑥)=4𝑥×4=44𝑥.cotcsccsc

Substituting these expressions, we have ddsincscsincsc𝑦𝑥=3(44𝑥)+344𝑥=124𝑥124𝑥.

Now, we have computed the derivatives of all three reciprocal functions and worked out examples using each formula. It may appear overwhelming at first to memorize these formulas, but there are useful patterns to keep in mind to help reduce mistakes. In particular, let us discuss an important pattern which rises from the cofunction identities. We begin by listing all six derivatives here: ddsincosddcossinddtansecddcotcscddsecsectanddcsccsccot𝑥𝑥=𝑥,𝑥𝑥=𝑥;𝑥𝑥=𝑥,𝑥𝑥=𝑥;𝑥𝑥=𝑥𝑥,𝑥𝑥=𝑥𝑥.

The three derivatives on the right sides are the cofunction derivatives corresponding to the ones on their left. For instance, cos𝑥 is a cofunction of sin𝑥, which means that cossin𝑥=𝜋2𝑥. From the list above, we can observe the following useful property.

Property: Derivatives of Cofunctions

Given the derivative of a trigonometric function, the derivative of its cofunction is obtained by multiplying the original derivative by 1 and interchanging each trigonometric function in the derivative with its respective cofunction.

For instance, if we know that (𝑥)=𝑥𝑥secsectan, then the derivative of the cofunction csc𝑥 is obtained by placing a negative sign in front of it and replacing sec𝑥 and tan𝑥 with their respective cofunctions csc𝑥 and cot𝑥, leading to (𝑥)=𝑥𝑥csccsccot.

Using this property, it is sufficient to only know three derivatives rather than six. By memorizing the derivatives of the sine, tangent, and secant functions, we can easily determine the derivatives of all six trigonometric and reciprocal trigonometric functions.

In our final example, we will use multiple formulas for the derivatives of reciprocal trigonometric functions to find a derivative.

Example 6: Finding the First Derivative of a Trigonometric Function Raised to a Negative Exponent

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

Answer

The function that we want to differentiate involves the cotangent and cosecant functions, so we can begin by recalling these derivatives: ddcotcscddcsccsccot𝑥𝑥=𝑥,𝑥𝑥=𝑥𝑥.

To find the derivative dd𝑦𝑥, we need to differentiate the function (75𝑥+36𝑥)cotcsc. We can approach this differentiation using two different methods. The first method is to apply the chain rule to this function, where the outside function is 𝑢 and the inside function is 75𝑥+36𝑥cotcsc. The second method is to rewrite this expression as a quotient and apply the quotient rule. We will choose the latter, writing this expression as a quotient. We need to compute ddddcotcsc𝑦𝑥=𝑥175𝑥+36𝑥.

Recall the quotient rule: given differentiable functions 𝑓(𝑥) and 𝑔(𝑥), ddif𝑥𝑓(𝑥)𝑔(𝑥)=𝑓(𝑥)𝑔(𝑥)𝑓(𝑥)𝑔(𝑥)(𝑔(𝑥))𝑔(𝑥)0.

Applying the quotient rule, we can write ddcotcsccotcsccotcsc𝑦𝑥=(1)(75𝑥+36𝑥)1(75𝑥+36𝑥)(75𝑥+36𝑥).

By the constant rule, we know that (1)=0. For the derivative (75𝑥+36𝑥)cotcsc, we can apply the sum/difference rule to separate the sum and also apply the constant multiple rule to factor out 7 and 3 in front of the derivatives: (75𝑥+36𝑥)=(75𝑥)+(36𝑥)=7(5𝑥)+3(6𝑥).cotcsccotcsccotcsc

Now we need to compute the derivatives (5𝑥)cot and (6𝑥)csc; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions 𝑓(𝑢) and 𝑔(𝑥), dd𝑥𝑓(𝑔(𝑥))=𝑓(𝑔(𝑥))𝑔(𝑥).

For cot5𝑥, the outside function is 𝑓(𝑢)=𝑢cot, while the inside function is 𝑔(𝑥)=5𝑥. Since we know 𝑓(𝑢)=𝑢csc and 𝑔(𝑥)=5 by the power rule, we can write the following using the chain rule: (5𝑥)=5𝑥×5=55𝑥.cotcsccsc

For csc6𝑥, the outside function is 𝑓(𝑢)=𝑢csc, while the inside function is 𝑔(𝑥)=6𝑥. Using 𝑓(𝑢)=𝑢𝑢csccot and 𝑔(𝑥)=6, (6𝑥)=6𝑥6𝑥×6=66𝑥6𝑥.csccsccotcsccot

Substituting these expressions, we have (75𝑥+36𝑥)=755𝑥+3(66𝑥6𝑥)=355𝑥186𝑥6𝑥.cotcsccsccsccotcsccsccot

Finally, we can substitute this expression in the quotient rule above to obtain ddcotcsccsccsccotcotcsccsccsccotcotcsccsccotcsccotcsc𝑦𝑥=0(75𝑥+36𝑥)1355𝑥186𝑥6𝑥(75𝑥+36𝑥)=355𝑥+186𝑥6𝑥(75𝑥+36𝑥)=186𝑥6𝑥+355𝑥(75𝑥+36𝑥).

Let us finish by recapping a few important concepts from the explainer.

Key Points

  • We can obtain the derivatives of reciprocal trigonometric functions by applying the quotient rule to the derivative rules for sine, cosine, and tangent functions.
  • Derivatives of the reciprocal trigonometric functions are ddsecsectanifforanyddcsccsccotifforanyddcotcscifforany𝑥𝑥=𝑥𝑥𝑥𝜋2+𝑛𝜋𝑛,𝑥𝑥=𝑥𝑥𝑥𝑛𝜋𝑛,𝑥𝑥=𝑥𝑥𝑛𝜋𝑛.
  • Given the derivative of a trigonometric function, the derivative of its cofunction is obtained by multiplying the original derivative by 1 and interchanging each trigonometric function in the derivative with its respective cofunction. This pattern can be observed in the following list of derivatives: ddsincosddcossinddtansecddcotcscddsecsectanddcsccsccot𝑥𝑥=𝑥,𝑥𝑥=𝑥;𝑥𝑥=𝑥,𝑥𝑥=𝑥;𝑥𝑥=𝑥𝑥,𝑥𝑥=𝑥𝑥.

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  • Interactive Sessions
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  • Realistic Exam Questions

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