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:

Note that if for any .

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:

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

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

### Rule: Quotient Rule

Given differentiable functions and ,

Applying the quotient rule to the right-hand side of equation (1), we have

By the constant rule, we know that . We also know that . Hence, the right-hand side of the equation above simplifies to

We can further simplify this expression using the definitions and . The right-hand side of the equation above can be written as

This leads to the derivative of the secant function.

### Rule: Derivative of the Secant Function

If for any ,

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 and ,

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 , determine the rate of change of when .

### Answer

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

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

To find the derivative , we need to differentiate the function . The constant can be factored outside of the derivative by the constant multiple rule, which means

We will apply the chain rule to take care of the expression inside the secant function. We recall the chain rule: given two differentiable functions and ,

We can see that the outside function of is , and the inside function is . Hence, and . We know that

Applying the power rule to , we obtain

Hence, applying the chain rule to leads to

Finally, remembering the constant in front of this derivative, we obtain

Now, we need to evaluate this expression at . Substituting this value into , we obtain

We note that ; hence, we can find an equivalent angle by subtracting radians from this angle:

This means that

Now, we see that is a special angle in the unit circle that has the following trigonometric ratios:

Since and for any angle , we have

Finally, substituting these values above, we obtain

Hence, the rate of change of at the given value is .

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 and ,

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 , find .

### Answer

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

To find the derivative , we need to differentiate the function that is a product of two functions. The constant can be factored outside of the derivative by the constant multiple rule, which leads to

Now, to differentiate , we notice that this is a product of two functions. Hence, we recall the product rule: given two differentiable functions and ,

Applying the product rule, we can write

Finally, we need to compute the derivatives and ; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions and ,

For , the outside function is , while the inside function is . Since we know and by the power rule, we can write the following using the chain rule:

For , the outside function is , while the inside function is . Using and ,

Substituting these expressions into the product rule above,

We must not forget the constant that we had factored outside of the derivative. Hence, we obtain

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

Applying the quotient rule,

By the constant rule, we know that . We also know that . Hence, the right-hand side of the equation above simplifies to

We can further simplify this expression using the definitions and . The right-hand side of the equation above can be written as

This leads to the derivative of the cosecant function.

### Rule: Derivative of the Cosecant Function

If for any ,

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 , find .

### Answer

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

To find the derivative , we need to differentiate the function that is a product of two functions. We recall the product rule: given two differentiable functions and ,

Applying the product rule, we can write

The first factor is the derivative of a polynomial, so we can compute this derivative by the power rule:

The last factor is the derivative of a function that is the sum of the cosecant function and a polynomial. Then,

Substituting these expressions into the product rule, we have

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 , find .

### Answer

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

To find the derivative , we need to differentiate the function . The constant can be factored outside of the derivative by the constant multiple rule, which means

Since is a composition of two functions, we recall the chain rule: given two differentiable functions and ,

We can see that the outside function of is , and the inside function is . Hence, . For the inside function , we note that by the constant rule, and by the power rule. This leads to . Substituting these expressions into the chain rule, we obtain

Finally, remembering the constant in front of this derivative, we obtain

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

Applying the quotient rule,

We know that

Hence, the derivative of can be written as

We can apply the trigonometric identity to write this derivative as . Finally, using the definition , the resulting expression can be written as . This leads to the derivative of the cotangent function.

### Rule: Derivative of the Cotangent Function

If for any ,

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 , given that .

### Answer

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

To find , we need to differentiate the function . The sum can be split up using the sum/difference rule, and the constants and 3 can be factored outside of each derivative leading to

Now we need to compute the derivatives and ; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions and ,

For , the outside function is , while the inside function is . Since we know and by the power rule, we can write the following using the chain rule:

For , the outside function is , while the inside function is . Using and ,

Substituting these expressions, we have

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:

The three derivatives on the right sides are the cofunction derivatives corresponding to the ones on their left. For instance, is a cofunction of , which means that . 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 and interchanging each trigonometric function in the derivative with its respective cofunction.

For instance, if we know that , then the derivative of the cofunction is obtained by placing a negative sign in front of it and replacing and with their respective cofunctions and , leading to .

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 , find .

### Answer

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

To find the derivative , we need to differentiate the function . 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 and the inside function is . 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

Recall the quotient rule: given differentiable functions and ,

Applying the quotient rule, we can write

By the constant rule, we know that . For the derivative , 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:

Now we need to compute the derivatives and ; both of these functions are compositions, which requires the chain rule. We recall the chain rule: given two differentiable functions and ,

For , the outside function is , while the inside function is . Since we know and by the power rule, we can write the following using the chain rule:

For , the outside function is , while the inside function is . Using and ,

Substituting these expressions, we have

Finally, we can substitute this expression in the quotient rule above to obtain

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
- Given the derivative of a trigonometric function, the derivative of its cofunction is obtained by multiplying the original derivative by and interchanging each trigonometric function in the derivative with its respective cofunction. This pattern can be observed in the following list of derivatives: