Lesson Explainer: Differentiation of Reciprocal Trigonometric Functions Mathematics • Higher Education

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: seccosifforanycscsinifforanycotcossinifforanyπ‘₯=1π‘₯π‘₯β‰ πœ‹2+π‘›πœ‹π‘›βˆˆβ„€,π‘₯=1π‘₯π‘₯β‰ π‘›πœ‹π‘›βˆˆβ„€,π‘₯=π‘₯π‘₯π‘₯β‰ π‘›πœ‹π‘›βˆˆβ„€.

Note that cottanπ‘₯=1π‘₯ if π‘₯β‰ πœ‹2+π‘›πœ‹ for any π‘›βˆˆβ„€.

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: ddsincosddcossinπ‘₯(π‘₯)=π‘₯,π‘₯(π‘₯)=βˆ’π‘₯.

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

ddsecddcosπ‘₯π‘₯=π‘₯ο€Ό1π‘₯.(1)

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

Rule: Quotient Rule

Given differentiable functions 𝑓(π‘₯) and 𝑔(π‘₯), ddifπ‘₯𝑓(π‘₯)𝑔(π‘₯)=𝑓′(π‘₯)𝑔(π‘₯)βˆ’π‘“(π‘₯)𝑔′(π‘₯)(𝑔(π‘₯))𝑔(π‘₯)β‰ 0.

Applying the quotient rule to the right-hand side of equation (1), we have ddcoscoscoscosπ‘₯ο€Ό1π‘₯=(1)β€²π‘₯βˆ’1(π‘₯)β€²(π‘₯).

By the constant rule, we know that (1)β€²=0. We also know that (π‘₯)β€²=βˆ’π‘₯cossin. Hence, the right-hand side of the equation above simplifies to 0π‘₯βˆ’1(βˆ’π‘₯)(π‘₯)=π‘₯π‘₯.cossincossincos

We can further simplify this expression using the definitions tansincosπ‘₯=π‘₯π‘₯ and seccosπ‘₯=1π‘₯. The right-hand side of the equation above can be written as sincoscostansecπ‘₯π‘₯Γ—1π‘₯=π‘₯π‘₯.

This leads to the derivative of the secant function.

Rule: Derivative of the Secant Function

If π‘₯β‰ πœ‹2+π‘›πœ‹ for any π‘›βˆˆβ„€, ddsecsectanπ‘₯π‘₯=π‘₯π‘₯.

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(𝑒) and 𝑔(π‘₯), ddπ‘₯(𝑓(𝑔(π‘₯)))=𝑓′(𝑔(π‘₯))𝑔′(π‘₯).

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=βˆ’22π‘₯sec, determine the rate of change of 𝑦 when π‘₯=11πœ‹6.

Answer

Recall that the rate of change of 𝑦 is given by the derivative dd𝑦π‘₯. In this example, we can find the rate of change of 𝑦 when π‘₯=11πœ‹6 by first finding the derivative dd𝑦π‘₯ and then evaluating the derivative at π‘₯=11πœ‹6.

The function that we want to differentiate involves the secant function. Hence, we begin by recalling the derivative of the secant function: ddsecsectanπ‘₯π‘₯=π‘₯π‘₯.

To find the derivative dd𝑦π‘₯, we need to differentiate the function βˆ’22π‘₯sec. The constant βˆ’2 can be factored outside of the derivative by the constant multiple rule, which means ddddsecddsec𝑦π‘₯=π‘₯(βˆ’22π‘₯)=βˆ’2π‘₯(2π‘₯).

We will apply the chain rule to take care of the expression 2π‘₯ inside the secant function. We recall the chain rule: given two differentiable functions 𝑓(𝑒) and 𝑔(π‘₯), ddπ‘₯𝑓(𝑔(π‘₯))=𝑓′(𝑔(π‘₯))𝑔′(π‘₯).

We can see that the outside function of sec2π‘₯ is sec𝑒, and the inside function is 2π‘₯. Hence, 𝑓(𝑒)=𝑒sec and 𝑔(π‘₯)=2π‘₯. We know that 𝑓′(𝑒)=𝑒𝑒=𝑒𝑒.ddsecsectan

Applying the power rule to 𝑔(π‘₯), we obtain 𝑔′(π‘₯)=π‘₯2π‘₯=2.dd

Hence, applying the chain rule to sec2π‘₯ leads to 𝑓′(𝑔(π‘₯))𝑔′(π‘₯)=2π‘₯2π‘₯Γ—2=22π‘₯2π‘₯.sectansectan

Finally, remembering the constant βˆ’2 in front of this derivative, we obtain ddsectansectan𝑦π‘₯=βˆ’2(22π‘₯2π‘₯)=βˆ’42π‘₯2π‘₯.

Now, we need to evaluate this expression at π‘₯=11πœ‹6. Substituting this value into dd𝑦π‘₯, we obtain ddsectan𝑦π‘₯|||=βˆ’411πœ‹311πœ‹3.ο—οŠ²οŽ οŽ ο‘½οŽ₯

We note that 11πœ‹3>2πœ‹; hence, we can find an equivalent angle by subtracting 2πœ‹ radians from this angle: 11πœ‹3βˆ’2πœ‹=11πœ‹3βˆ’6πœ‹3=5πœ‹3.

This means that ddsectan𝑦π‘₯|||=βˆ’45πœ‹35πœ‹3.ο—οŠ²οŽ οŽ ο‘½οŽ₯

Now, we see that 5πœ‹3 is a special angle in the unit circle that has the following trigonometric ratios: sincos5πœ‹3=βˆ’βˆš32,5πœ‹3=12.

Since secsinπœƒ=1πœƒ and tansincosπœƒ=πœƒπœƒ for any angle πœƒ, we have sectan5πœ‹3=1=2,5πœ‹3==βˆ’βˆš3.√

Finally, substituting these values above, we obtain dd𝑦π‘₯|||=βˆ’4Γ—2Γ—ο€»βˆ’βˆš3=8√3.ο—οŠ²οŽ οŽ ο‘½οŽ₯

Hence, the rate of change of 𝑦 at the given π‘₯ value is 8√3.

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(π‘₯) and 𝑔(π‘₯), ddπ‘₯(𝑓(π‘₯)𝑔(π‘₯))=𝑓′(π‘₯)𝑔(π‘₯)+𝑓(π‘₯)𝑔′(π‘₯).

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=βˆ’98π‘₯8π‘₯tansec, find dd𝑦π‘₯.

Answer

The function that we want to differentiate involves the tangent and secant functions, so we can begin by recalling these derivatives: ddtansecddsecsectanπ‘₯π‘₯=π‘₯,π‘₯π‘₯=π‘₯π‘₯.

To find the derivative dd𝑦π‘₯, we need to differentiate the function βˆ’98π‘₯8π‘₯tansec that is a product of two functions. The constant βˆ’9 can be factored outside of the derivative by the constant multiple rule, which leads to ddddtansecddtansec𝑦π‘₯=π‘₯(βˆ’98π‘₯8π‘₯)=βˆ’9π‘₯(8π‘₯8π‘₯).

Now, to differentiate tansec8π‘₯8π‘₯, we notice that this is a product of two functions. Hence, we recall the product rule: given two differentiable functions 𝑓(π‘₯) and 𝑔(π‘₯), ddπ‘₯(𝑓(π‘₯)𝑔(π‘₯))=𝑓′(π‘₯)𝑔(π‘₯)+𝑓(π‘₯)𝑔′(π‘₯).

Applying the product rule, we can write ddtansectansectansecπ‘₯(8π‘₯8π‘₯)=(8π‘₯)β€²8π‘₯+8π‘₯(8π‘₯)β€².

Finally, we need to compute the derivatives (8π‘₯)β€²tan and (8π‘₯)β€²sec; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions 𝑓(𝑒) and 𝑔(π‘₯), ddπ‘₯𝑓(𝑔(π‘₯))=𝑓′(𝑔(π‘₯))𝑔′(π‘₯).

For tan8π‘₯, the outside function is 𝑓(𝑒)=𝑒tan, while the inside function is 𝑔(π‘₯)=8π‘₯. Since we know 𝑓′(𝑒)=𝑒sec and 𝑔′(π‘₯)=8 by the power rule, we can write the following using the chain rule: (8π‘₯)β€²=8π‘₯Γ—8=88π‘₯.tansecsec

For sec8π‘₯, the outside function is 𝑓(𝑒)=𝑒sec, while the inside function is 𝑔(π‘₯)=8π‘₯. Using 𝑓′(𝑒)=𝑒𝑒sectan and 𝑔′(π‘₯)=8, (8π‘₯)β€²=8π‘₯8π‘₯Γ—8=88π‘₯8π‘₯.secsectansectan

Substituting these expressions into the product rule above, ddtansecsecsectansectansecsectanπ‘₯(8π‘₯8π‘₯)=88π‘₯8π‘₯+8π‘₯(88π‘₯8π‘₯)=88π‘₯+88π‘₯8π‘₯.

We must not forget the constant βˆ’9 that we had factored outside of the derivative. Hence, we obtain ddsecsectansecsectansectansec𝑦π‘₯=βˆ’9ο€Ή88π‘₯+88π‘₯8π‘₯=βˆ’728π‘₯βˆ’728π‘₯8π‘₯=βˆ’728π‘₯8π‘₯βˆ’728π‘₯.

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 ddcscddsinπ‘₯π‘₯=π‘₯ο€Ό1π‘₯.

Applying the quotient rule, ddsinsinsinsinπ‘₯ο€Ό1π‘₯=(1)β€²π‘₯βˆ’1(π‘₯)β€²(π‘₯).

By the constant rule, we know that (1)β€²=0. We also know that (π‘₯)β€²=π‘₯sincos. Hence, the right-hand side of the equation above simplifies to 0π‘₯βˆ’1(π‘₯)(π‘₯)=βˆ’π‘₯π‘₯.sincossincossin

We can further simplify this expression using the definitions cotcossinπ‘₯=π‘₯π‘₯ and cscsinπ‘₯=1π‘₯. The right-hand side of the equation above can be written as βˆ’π‘₯π‘₯Γ—1π‘₯=βˆ’π‘₯π‘₯.cossinsincotcsc

This leads to the derivative of the cosecant function.

Rule: Derivative of the Cosecant Function

If π‘₯β‰ π‘›πœ‹ for any π‘›βˆˆβ„€, ddcsccsccotπ‘₯π‘₯=βˆ’π‘₯π‘₯.

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 InvolvingReciprocal TrigonometricRatios Using the Product Rule

Given 𝑦=(π‘₯+3)(9π‘₯+π‘₯)csc, find dd𝑦π‘₯.

Answer

The function that we want to differentiate involves the cosecant function, so we can begin by recalling the derivative of the cosecant function: ddcsccsccotπ‘₯π‘₯=βˆ’π‘₯π‘₯.

To find the derivative dd𝑦π‘₯, we need to differentiate the function (π‘₯+3)(9π‘₯+π‘₯)csc that is a product of two functions. We recall the product rule: given two differentiable functions 𝑓(π‘₯) and 𝑔(π‘₯), ddπ‘₯(𝑓(π‘₯)𝑔(π‘₯))=𝑓′(π‘₯)𝑔(π‘₯)+𝑓(π‘₯)𝑔′(π‘₯).

Applying the product rule, we can write ddddcsccsccsc𝑦π‘₯=π‘₯((π‘₯+3)(9π‘₯+π‘₯))=(π‘₯+3)β€²(9π‘₯+π‘₯)+(π‘₯+3)(9π‘₯+π‘₯)β€².

The first factor (π‘₯+3)β€² is the derivative of a polynomial, so we can compute this derivative by the power rule: (π‘₯+3)β€²=1+0=1.

The last factor (9π‘₯+π‘₯)β€²csc is the derivative of a function that is the sum of the cosecant function and a polynomial. Then, (9π‘₯+π‘₯)β€²=(9π‘₯)β€²+(π‘₯)β€²=9βˆ’π‘₯π‘₯.csccsccsccot

Substituting these expressions into the product rule, we have ddcsccsccotcsccsccotcsccotcsc𝑦π‘₯=1(9π‘₯+π‘₯)+(π‘₯+3)(9βˆ’π‘₯π‘₯)=9π‘₯+π‘₯+9π‘₯+27βˆ’(π‘₯+3)π‘₯π‘₯=18π‘₯βˆ’(π‘₯+3)π‘₯π‘₯+π‘₯+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=βˆ’13(πœ‹+5π‘₯)csc, find dd𝑦π‘₯.

Answer

The function that we want to differentiate involves the cosecant function. Hence, we begin by recalling the derivative of the secant function: ddcsccsccotπ‘₯π‘₯=βˆ’π‘₯π‘₯.

To find the derivative dd𝑦π‘₯, we need to differentiate the function βˆ’13(πœ‹+5π‘₯)csc. The constant βˆ’13 can be factored outside of the derivative by the constant multiple rule, which means ddddcscddcsc𝑦π‘₯=π‘₯(βˆ’13(πœ‹+5π‘₯))=βˆ’13π‘₯(πœ‹+5π‘₯).

Since csc(πœ‹+5π‘₯) is a composition of two functions, we recall the chain rule: given two differentiable functions 𝑓(𝑒) and 𝑔(π‘₯), ddπ‘₯𝑓(𝑔(π‘₯))=𝑓′(𝑔(π‘₯))𝑔′(π‘₯).

We can see that the outside function of csc(πœ‹+5π‘₯) is 𝑓(𝑒)=𝑒csc, and the inside function is 𝑔(π‘₯)=πœ‹+5π‘₯. Hence, 𝑓′(𝑒)=βˆ’π‘’π‘’csccot. For the inside function 𝑔, we note that (πœ‹)β€²=0 by the constant rule, and (5π‘₯)β€²=5 by the power rule. This leads to 𝑔′(π‘₯)=5. Substituting these expressions into the chain rule, we obtain ddcsccsccotcsccotπ‘₯(πœ‹+5π‘₯)=βˆ’(πœ‹+5π‘₯)(πœ‹+5π‘₯)Γ—5=βˆ’5(πœ‹+5π‘₯)(πœ‹+5π‘₯).

Finally, remembering the constant βˆ’13 in front of this derivative, we obtain ddcsccotcsccot𝑦π‘₯=βˆ’13(βˆ’5(πœ‹+5π‘₯)(πœ‹+5π‘₯))=65(πœ‹+5π‘₯)(πœ‹+5π‘₯).

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 cosecant function, we can write ddcotddcossinπ‘₯π‘₯=π‘₯ο€»π‘₯π‘₯.

Applying the quotient rule, ddcossincossincossinsinπ‘₯ο€»π‘₯π‘₯=(π‘₯)β€²π‘₯βˆ’π‘₯(π‘₯)β€²(π‘₯).

We know that (π‘₯)β€²=βˆ’π‘₯,(π‘₯)β€²=π‘₯.cossinsincos

Hence, the derivative of cotπ‘₯ can be written as βˆ’π‘₯Γ—π‘₯βˆ’π‘₯Γ—π‘₯(π‘₯)=βˆ’ο€Ίπ‘₯+π‘₯π‘₯.sinsincoscossinsincossin

We can apply the trigonometric identity sincosπ‘₯+π‘₯=1 to write this derivative as βˆ’1π‘₯sin. Finally, using the definition cscsinπ‘₯=1π‘₯, the resulting expression can be written as βˆ’π‘₯csc. This leads to the derivative of the cotangent function.

Rule: Derivative of the Cotangent Function

If π‘₯β‰ π‘›πœ‹ for any π‘›βˆˆβ„€, ddcotcscπ‘₯π‘₯=βˆ’π‘₯.

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 dd𝑦π‘₯, given that 𝑦=βˆ’34π‘₯+34π‘₯coscot.

Answer

The function that we want to differentiate involves the cosine and cotangent functions, so we can begin by recalling these derivatives: ddcossinddcotcscπ‘₯π‘₯=βˆ’π‘₯,π‘₯π‘₯=βˆ’π‘₯.

To find dd𝑦π‘₯, we need to differentiate the function βˆ’34π‘₯+34π‘₯coscot. The sum can be split up using the sum/difference rule, and the constants βˆ’3 and 3 can be factored outside of each derivative leading to ddddcoscotddcosddcotddcosddcot𝑦π‘₯=π‘₯(βˆ’34π‘₯+34π‘₯)=π‘₯(βˆ’34π‘₯)+π‘₯(34π‘₯)=βˆ’3π‘₯(4π‘₯)+3π‘₯(4π‘₯).

Now we need to compute the derivatives (4π‘₯)β€²cos and (4π‘₯)β€²cot; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions 𝑓(𝑒) and 𝑔(π‘₯), ddπ‘₯𝑓(𝑔(π‘₯))=𝑓′(𝑔(π‘₯))𝑔′(π‘₯).

For cos4π‘₯, the outside function is 𝑓(𝑒)=𝑒cos, while the inside function is 𝑔(π‘₯)=4π‘₯. Since we know 𝑓′(𝑒)=βˆ’π‘’sin and 𝑔′(π‘₯)=4 by the power rule, we can write the following using the chain rule: (4π‘₯)β€²=βˆ’4π‘₯Γ—4=βˆ’44π‘₯.cossinsin

For cot4π‘₯, the outside function is 𝑓(𝑒)=𝑒cot, while the inside function is 𝑔(π‘₯)=4π‘₯. Using 𝑓′(𝑒)=βˆ’π‘’csc and 𝑔′(π‘₯)=4, (4π‘₯)β€²=βˆ’4π‘₯Γ—4=βˆ’44π‘₯.cotcsccsc

Substituting these expressions, we have ddsincscsincsc𝑦π‘₯=βˆ’3(βˆ’44π‘₯)+3ο€Ήβˆ’44π‘₯=124π‘₯βˆ’124π‘₯.

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: ddsincosddcossinddtansecddcotcscddsecsectanddcsccsccotπ‘₯π‘₯=π‘₯,π‘₯π‘₯=βˆ’π‘₯;π‘₯π‘₯=π‘₯,π‘₯π‘₯=βˆ’π‘₯;π‘₯π‘₯=π‘₯π‘₯,π‘₯π‘₯=βˆ’π‘₯π‘₯.

The three derivatives on the right sides are the cofunction derivatives corresponding to the ones on their left. For instance, cosπ‘₯ is a cofunction of sinπ‘₯, which means that cossinπ‘₯=ο€»πœ‹2βˆ’π‘₯. 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 βˆ’1 and interchanging each trigonometric function in the derivative with its respective cofunction.

For instance, if we know that (π‘₯)β€²=π‘₯π‘₯secsectan, then the derivative of the cofunction cscπ‘₯ is obtained by placing a negative sign in front of it and replacing secπ‘₯ and tanπ‘₯ with their respective cofunctions cscπ‘₯ and cotπ‘₯, leading to (π‘₯)β€²=βˆ’π‘₯π‘₯csccsccot.

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=(75π‘₯+36π‘₯)cotcsc, find dd𝑦π‘₯.

Answer

The function that we want to differentiate involves the cotangent and cosecant functions, so we can begin by recalling these derivatives: ddcotcscddcsccsccotπ‘₯π‘₯=βˆ’π‘₯,π‘₯π‘₯=βˆ’π‘₯π‘₯.

To find the derivative dd𝑦π‘₯, we need to differentiate the function (75π‘₯+36π‘₯)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 π‘’οŠ±οŠ§ and the inside function is 75π‘₯+36π‘₯cotcsc. 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 ddddcotcsc𝑦π‘₯=π‘₯ο€Ό175π‘₯+36π‘₯.

Recall the quotient rule: given differentiable functions 𝑓(π‘₯) and 𝑔(π‘₯), ddifπ‘₯𝑓(π‘₯)𝑔(π‘₯)=𝑓′(π‘₯)𝑔(π‘₯)βˆ’π‘“(π‘₯)𝑔′(π‘₯)(𝑔(π‘₯))𝑔(π‘₯)β‰ 0.

Applying the quotient rule, we can write ddcotcsccotcsccotcsc𝑦π‘₯=(1)β€²(75π‘₯+36π‘₯)βˆ’1(75π‘₯+36π‘₯)β€²(75π‘₯+36π‘₯).

By the constant rule, we know that (1)β€²=0. For the derivative (75π‘₯+36π‘₯)β€²cotcsc, 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: (75π‘₯+36π‘₯)β€²=(75π‘₯)β€²+(36π‘₯)β€²=7(5π‘₯)β€²+3(6π‘₯)β€².cotcsccotcsccotcsc

Now we need to compute the derivatives (5π‘₯)β€²cot and (6π‘₯)β€²csc; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions 𝑓(𝑒) and 𝑔(π‘₯), ddπ‘₯𝑓(𝑔(π‘₯))=𝑓′(𝑔(π‘₯))𝑔′(π‘₯).

For cot5π‘₯, the outside function is 𝑓(𝑒)=𝑒cot, while the inside function is 𝑔(π‘₯)=5π‘₯. Since we know 𝑓′(𝑒)=βˆ’π‘’csc and 𝑔′(π‘₯)=5 by the power rule, we can write the following using the chain rule: (5π‘₯)β€²=βˆ’5π‘₯Γ—5=βˆ’55π‘₯.cotcsccsc

For csc6π‘₯, the outside function is 𝑓(𝑒)=𝑒csc, while the inside function is 𝑔(π‘₯)=6π‘₯. Using 𝑓′(𝑒)=βˆ’π‘’π‘’csccot and 𝑔′(π‘₯)=6, (6π‘₯)β€²=βˆ’6π‘₯6π‘₯Γ—6=βˆ’66π‘₯6π‘₯.csccsccotcsccot

Substituting these expressions, we have (75π‘₯+36π‘₯)β€²=7ο€Ήβˆ’55π‘₯+3(βˆ’66π‘₯6π‘₯)=βˆ’355π‘₯βˆ’186π‘₯6π‘₯.cotcsccsccsccotcsccsccot

Finally, we can substitute this expression in the quotient rule above to obtain ddcotcsccsccsccotcotcsccsccsccotcotcsccsccotcsccotcsc𝑦π‘₯=0(75π‘₯+36π‘₯)βˆ’1ο€Ήβˆ’355π‘₯βˆ’186π‘₯6π‘₯(75π‘₯+36π‘₯)=355π‘₯+186π‘₯6π‘₯(75π‘₯+36π‘₯)=186π‘₯6π‘₯+355π‘₯(75π‘₯+36π‘₯).

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 ddsecsectanifforanyddcsccsccotifforanyddcotcscifforanyπ‘₯π‘₯=π‘₯π‘₯π‘₯β‰ πœ‹2+π‘›πœ‹π‘›βˆˆβ„€,π‘₯π‘₯=βˆ’π‘₯π‘₯π‘₯β‰ π‘›πœ‹π‘›βˆˆβ„€,π‘₯π‘₯=βˆ’π‘₯π‘₯β‰ π‘›πœ‹π‘›βˆˆβ„€.
  • Given the derivative of a trigonometric function, the derivative of its cofunction is obtained by multiplying the original derivative by βˆ’1 and interchanging each trigonometric function in the derivative with its respective cofunction. This pattern can be observed in the following list of derivatives: ddsincosddcossinddtansecddcotcscddsecsectanddcsccsccotπ‘₯π‘₯=π‘₯,π‘₯π‘₯=βˆ’π‘₯;π‘₯π‘₯=π‘₯,π‘₯π‘₯=βˆ’π‘₯;π‘₯π‘₯=π‘₯π‘₯,π‘₯π‘₯=βˆ’π‘₯π‘₯.

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