# 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:

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 and , the derivative of their product is given by

This can be written succinctly using prime notation as follows:

### Quotient Rule

Given two differentiable functions and , the derivative of their quotient is given by

This can be written succinctly using prime notation as follows:

### Example 1: Derivative of the Secant Function

If , determine the rate of change of when .

To find the rate of change of at , we will first find the derivative and evaluate it at this point.

Using the definition of the secant function,

we can rewrite the expression for as

We can now apply the quotient rule, to . Setting and , we have

Substituting these expressions into the quotient rule, we have

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

Hence, substituting in into our expression gives

Using the technique we used in the last example, it is possible to prove the general rule for the derivative of the secant function:

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

### Example 2: Derivative of the Cotangent Function

Determine given that .

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, 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 hence, we can write

We can now apply the quotient rule, by setting and . We can now differentiate and , using the standard results for the derivatives of sine and cosine to get

Substituting the expressions for , , , and into the quotient rule gives

Using the Pythagorean identity for sine and cosine, we can simplify the numerator to get

Substituting this equation and equation (2) into equation (1), we get

In the last example, we showed that

Using the same techniques, we can generalize the result to

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

### Example 3: Derivative of the Cosecant Function

Given that , find .

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 . Using the definition of the cosecant function, we have

To differentiate this, we can apply the quotient rule:

Setting and , we have

Therefore,

Now we know the derivative of , we can consider the effect of adding to .

Consider the graph of a general function . What is the effect of plotting for some constant ? The graph of will be the graph of translated by 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 at a point will be equal to the derivative of at . Hence,

Therefore, given that the derivative of is , we can conclude that

Hence,

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:

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 , find .

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:

Setting and , we have

Substituting these expressions into the product rule, we have

### Example 5: Differentiating Trigonometric Functions

If , find .

To find this derivative, we will apply the product rule:

Setting and , we have

Substituting these expressions into the product rule, we have

The function in the previous example is actually the derivative of . 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 , find .

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:

Setting and , we have

Substituting these expressions into the quotient rule, we have

### 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:
3. Using these standard results in conjunction with the rules of differentiation, we can find the derivatives of a large class of functions.