# Lesson Explainer: Derivatives of Inverse Trigonometric Functions Mathematics

In this explainer, we will learn how to find the derivatives of the inverses of trigonometric functions.

When we consider the function , we are giving an input of the measure of an angle in radians and getting the sine of that angle. Oftentimes, we might want to know what measure of angle results in a given value of sine, say . In this case, we write or

In this explainer, we will use the inverse notation , although we need to be a little careful about how we define the inverse of a function like sine because sine is a periodic function. In fact, for any given sine, there are infinitely many real numbers that map to that value of sine.

Therefore, to define the inverse of sine, or any other trigonometric function for that matter, we need to restrict the domain of the function to ensure the inverse is single valued. In the case of sine, we restrict ourselves to .

In this explainer, we will be interested in computing the derivatives of the inverse trigonometric functions. To differentiate such function, we will need to use implicit differentiation, which, for single-variable functions, is a corollary of the chain rule. Below is a summary of the chain rule.

### Chain Rule

Let be a function that is differentiable at a point and let be a function that is differentiable at the point ; then, their composite function is differentiable at and

If we use the notation that and , then the chain rule states

In the first example, we will use implicit differentiation to find the derivative of the inverse sine function.

### Example 1: Differentiating Inverse Sine Using Implicit Differentiation

Find .

We begin by setting . Applying sine to both sides, we have

We can now differentiate this equation with respect to using implicit differentiation. Taking the derivative of both sides of the equation, we have

On the right-hand side, the derivative of is simply the constant 1. As for the left-hand side, we can use the chain rule to differentiate with respect to as follows:

Hence,

We can divide both sides of the equation by when it is nonzero; that is, when . This gives us

 ddcos𝑦𝑥=1𝑦. (1)

This equation gives us an expression for the derivative in terms of , whereas we would like an expression in terms of . We know that ; therefore, if we express in terms of , we can change this to an expression in terms of . The Pythagorean identity for sine and cosine provides us with the necessary means of expressing in terms of . Recall that

Subtracting from both sides and taking the square root, we have

Since the range of inverse sine is , we have . Hence, within this range. Therefore, we need to take the positive square root as follows:

Substituting this into equation (1) gives us

Since , we can rewrite this in terms of as

Hence,

This formula is valid for values of in the range .

It is often helpful to visualize a graph and its derivative. Below is a plot of and its derivative.

By considering this graph and the graph of , we might guess that the derivative of inverse cosine will be equal to negative the derivative of inverse sine. As we shall see in the next example, this is the case.

In the next example, we will use a similar, but slightly different, technique to derive the formula for the derivative of the inverse cosine function. Instead of using implicit differentiation, like we did in the last example, we will use the inverse function theorem which we state below.

### Inverse function Theorem

Let be a differentiable function with a continuous derivative . If for some in the domain of , then is invertible in a neighborhood of and has a differentiable inverse such that

This is sometimes written simply as

### Example 2: Differentiating Inverse Cosine Using the Inverse Function Theorem

Find , where .

We begin by setting . Hence,

Multiplying through by , we get the following expression for in terms of :

We can now differentiate with respect to to get

If , . Therefore, we can apply the inverse function theorem to get

 ddsin𝑦𝑥=1=−1𝑎𝑦.dd (2)

We now have an expression for the derivative in terms of . However, we would like to express this in terms of . We can use the Pythagorean identity for sine and cosine, to express in terms of as follows:

Since when , we have

We can now substitute this expression for into equation (2):

Since , we can rewrite this as

Hence, for .

The figure below shows the graph of and its derivative.

### Example 3: Differentiating Inverse Tangent

Find an expression for the derivative of in terms of .

Given that , we can write

We can now (implicitly) differentiate both sides of the equation with respect to to get

Since for all values of , we have

We would like to rewrite the expression for the derivate in terms of . To do this, we would like to convert into an expression in terms of which we can then rewrite in terms of . We can use the Pythagorean identity for secant and tangent, to rewrite this as

Since , we have

In a similar way, it is possible to derive the formula from the derivative of the inverse cotangent function:

### Example 4: Differentiating Inverse Cosecant

Find .

We begin by setting . Hence, we can rewrite this as

Differentiating both sides with respect to , we have

Given that when and , we can write

 ddcsccot𝑦𝑥=−1𝑦𝑦. (3)

We would like to express this in terms of . To do this, we will appeal to the Pythagorean identity for the cosecant and cotangent functions:

Hence,

Substituting this into equation (3), we have

Rewriting this in terms of , we have

At this point, we need to confirm the sign of the derivative. By considering the graph of the inverse cosecant function, we see that, for all values of in the range of the function, the derivative is negative.

Therefore, we can use the absolute value to ensure that the derivative is always negative as follows:

This formula can be generalized to include a constant :

Using a similar technique, it is also possible to derive the formula for the inverse secant function:

We can summarize our results thus far as follows.

### Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions are as follows:

In many circumstances, it is possible to quote these results without derivation. Having said this, knowing how to derive these formulas is of great importance. Additionally, it will greatly benefit the student to commit results such as these to memory. Being able to easily recall results like this will enable you to tackle differentiation problems with confidence and efficiency. To aid with committing these to memory, it is helpful to note that the derivatives of each pair of cofunctions have equal but opposite derivatives. For example,

In the last couple of examples, we will apply these results to solve some additional problems on differentiation of inverse trigonometric functions.

### Example 5: Differentiating Inverse Trigonometric Functions

Evaluate .

Since we have a composition of functions, we will apply the chain rule. Remember, the chain rule states that, given and ,

Setting and , we have

Substituting these expressions into the formula for the chain rule, we have

Notice that the last example showed us that

This is, in fact, no accident, and

Therefore, if you were familiar with this identity, you could have solved the last example much more efficiently by simply applying the identity and then quoting the result about the derivative of the inverse tangent function.

### Example 6: Differentiating Inverse Trigonometric Functions

Evaluate .

We have been asked to differentiate the compositions of two functions. Therefore, we should use the chain rule, which states that, given and ,

Firstly, we need to identify our . In this case, there is really only one sensible choice, and that is the expression that the inverse sine function is being applied to. Hence, we set and . We now differentiate each of these. Beginning with , we have

Substituting in , we have

We now differentiate with respect to . To do this, we will need to use the chain rule again. This time we will set and . Differentiating both and , we have

Using the chain rule, we have

We now have an expression for and . Hence, we can apply the chain rule to find an expression for as follows:

Once again, we have an interesting result that

In a similar way to our last example, this result actually comes from the identity for . Familiarity with this result would have significantly reduced our computation in the previous example.

### Key Points

• Using implicit differentiation or the inverse function theorem, it is possible to derive the formulas for the derivatives of inverse trigonometric functions.
• The derivatives of the inverse trigonometric functions are as follows:
• Familiarity with trigonometric identities can oftentimes significantly simplify the process of finding derivatives.