Lesson Explainer: Derivatives of Inverse Trigonometric Functions Mathematics • Higher Education

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

When we consider the function 𝑦=π‘₯sin, 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 π‘₯ results in a given value of sine, say 𝑦. In this case, we write π‘₯=𝑦sin or π‘₯=𝑦.arcsin

In this explainer, we will use the inverse notation βˆ’1, 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 βˆ’πœ‹2≀π‘₯β‰€πœ‹2.

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 𝑔 be a function that is differentiable at a point π‘Ž and let 𝑓 be a function that is differentiable at the point 𝑔(π‘Ž); then, their composite function π‘“βˆ˜π‘” is differentiable at π‘Ž and (π‘“βˆ˜π‘”)(π‘Ž)=𝑓(𝑔(π‘Ž))𝑔(π‘Ž).

If we use the notation that 𝑦=𝑓(𝑒) and 𝑒=𝑔(π‘₯), then the chain rule states dddddd𝑦π‘₯=𝑦𝑒×𝑒π‘₯.

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 ddsinπ‘₯π‘₯.

Answer

We begin by setting 𝑦=π‘₯sin. Applying sine to both sides, we have sin𝑦=π‘₯.

We can now differentiate this equation with respect to π‘₯ using implicit differentiation. Taking the derivative of both sides of the equation, we have ddsinddπ‘₯(𝑦)=π‘₯(π‘₯).

On the right-hand side, the derivative of π‘₯ is simply the constant 1. As for the left-hand side, we can use the chain rule to differentiate sin𝑦 with respect to π‘₯ as follows: ddsindd𝑦(𝑦)𝑦π‘₯=1.

Hence, cosdd𝑦𝑦π‘₯=1.

We can divide both sides of the equation by cos𝑦 when it is nonzero; that is, when π‘¦β‰ πœ‹2,βˆ’πœ‹2. This gives us

ddcos𝑦π‘₯=1𝑦.(1)

This equation gives us an expression for the derivative in terms of 𝑦, whereas we would like an expression in terms of π‘₯. We know that sin𝑦=π‘₯; therefore, if we express cos𝑦 in terms of sin𝑦, we can change this to an expression in terms of π‘₯. The Pythagorean identity for sine and cosine provides us with the necessary means of expressing cos𝑦 in terms of sin𝑦. Recall that sincosοŠ¨οŠ¨π‘¦+𝑦=1.

Subtracting sinοŠ¨π‘¦ from both sides and taking the square root, we have cossin𝑦=±1βˆ’π‘¦.

Since the range of inverse sine is ο“βˆ’πœ‹2,πœ‹2, we have βˆ’πœ‹2β‰€π‘¦β‰€πœ‹2. Hence, 0≀𝑦≀1cos within this range. Therefore, we need to take the positive square root as follows: cossin𝑦=1βˆ’π‘¦.

Substituting this into equation (1) gives us ddsin𝑦π‘₯=11βˆ’π‘¦.

Since sin𝑦=π‘₯, we can rewrite this in terms of π‘₯ as dd𝑦π‘₯=1√1βˆ’π‘₯.

Hence, ddsinπ‘₯π‘₯=1√1βˆ’π‘₯.

This formula is valid for values of π‘₯ in the range βˆ’1<π‘₯<1.

It is often helpful to visualize a graph and its derivative. Below is a plot of 𝑦=π‘₯sin and its derivative.

By considering this graph and the graph of 𝑦=π‘₯cos, 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 𝑓 be a differentiable function with a continuous derivative π‘“οŽ˜. If 𝑓(π‘Ž)β‰ 0 for some π‘Ž in the domain of 𝑓, then 𝑓 is invertible in a neighborhood of π‘Ž and has a differentiable inverse such that 𝑓=1𝑓(π‘Ž).

This is sometimes written simply as ddπ‘₯𝑦=1.ddο˜ο—

Example 2: Differentiating Inverse Cosine Using the Inverse Function Theorem

Find ddcosπ‘₯ο€»π‘₯π‘Žο‡οŠ±οŠ§, where π‘Žβ‰ 0.

Answer

We begin by setting 𝑦=ο€»π‘₯π‘Žο‡cos. Hence, cos𝑦=π‘₯π‘Ž.

Multiplying through by π‘Ž, we get the following expression for π‘₯ in terms of 𝑦: π‘₯=π‘Žπ‘¦.cos

We can now differentiate with respect to 𝑦 to get ddsinπ‘₯𝑦=βˆ’π‘Žπ‘¦.

If 0<𝑦<πœ‹, sin𝑦≠0. Therefore, we can apply the inverse function theorem ddπ‘₯𝑦=1ddο˜ο— to get

ddsin𝑦π‘₯=1=βˆ’1π‘Žπ‘¦.ddο—ο˜(2)

We now have an expression for the derivative in terms of 𝑦. However, we would like to express this in terms of π‘₯. We can use the Pythagorean identity for sine and cosine, sincosοŠ¨οŠ¨π‘¦+𝑦=1, to express sin𝑦 in terms of cos𝑦 as follows: sincos𝑦=±√1βˆ’π‘¦.

Since sin𝑦>0 when 0<𝑦<πœ‹, we have sincos𝑦=√1βˆ’π‘¦.

We can now substitute this expression for sin𝑦 into equation (2): ddcos𝑦π‘₯=βˆ’1π‘Žβˆš1βˆ’π‘¦.

Since cos𝑦=π‘₯π‘Ž, we can rewrite this as dd𝑦π‘₯=βˆ’1π‘Žο„ž1βˆ’ο€»ο‡=βˆ’1βˆšπ‘Žβˆ’π‘₯.ο—οŒΊοŠ¨οŠ¨οŠ¨

Hence, ddcosπ‘₯ο€»π‘₯π‘Žο‡=βˆ’1βˆšπ‘Žβˆ’π‘₯, for βˆ’π‘Ž<π‘₯<π‘Ž.

The figure below shows the graph of 𝑦=ο€»π‘₯π‘Žο‡cos and its derivative.

Example 3: Differentiating Inverse Tangent

Find an expression for the derivative of 𝑦=π‘Žπ‘₯tan in terms of π‘₯.

Answer

Given that 𝑦=π‘Žπ‘₯tan, we can write tan𝑦=π‘Žπ‘₯.

We can now (implicitly) differentiate both sides of the equation with respect to π‘₯ to get secddοŠ¨π‘¦π‘¦π‘₯=π‘Ž.

Since secοŠ¨π‘¦β‰ 0 for all values of 𝑦, we have ddsec𝑦π‘₯=π‘Žπ‘¦.

We would like to rewrite the expression for the derivate in terms of π‘₯. To do this, we would like to convert secοŠ¨π‘¦ into an expression in terms of tan𝑦 which we can then rewrite in terms of π‘₯. We can use the Pythagorean identity for secant and tangent, 1+𝑦=𝑦,tansec to rewrite this as ddtan𝑦π‘₯=π‘Ž1+𝑦.

Since tan𝑦=π‘Žπ‘₯, we have dd𝑦π‘₯=π‘Ž1+(π‘Žπ‘₯).

In a similar way, it is possible to derive the formula from the derivative of the inverse cotangent function: ddcotπ‘₯ο€Ήπ‘Žπ‘₯=βˆ’π‘Ž1+(π‘Žπ‘₯).

Example 4: Differentiating Inverse Cosecant

Find ddcscπ‘₯π‘₯.

Answer

We begin by setting 𝑦=π‘₯csc. Hence, we can rewrite this as csc𝑦=π‘₯.

Differentiating both sides with respect to π‘₯, we have βˆ’π‘¦π‘¦π‘¦π‘₯=1.csccotdd

Given that csccot𝑦𝑦≠0 when βˆ’πœ‹2<𝑦<πœ‹2 and 𝑦≠0, we can write

ddcsccot𝑦π‘₯=βˆ’1𝑦𝑦.(3)

We would like to express this in terms of π‘₯. To do this, we will appeal to the Pythagorean identity for the cosecant and cotangent functions: cotcscοŠ¨οŠ¨π‘¦+1=𝑦.

Hence, cotcsc𝑦=Β±βˆšπ‘¦βˆ’1.

Substituting this into equation (3), we have ddcsccsc𝑦π‘₯=βˆ’1π‘¦ο€ΊΒ±βˆšπ‘¦βˆ’1.

Rewriting this in terms of π‘₯, we have dd𝑦π‘₯=βˆ’1π‘₯ο€»Β±βˆšπ‘₯βˆ’1.

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 π‘₯ 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: dd𝑦π‘₯=βˆ’|||1π‘₯√π‘₯βˆ’1|||=βˆ’1|π‘₯|√π‘₯βˆ’1.

This formula can be generalized to include a constant π‘Ž: ddcscπ‘₯ο€Ήπ‘Žπ‘₯=βˆ’1π‘Ž|π‘₯|√(π‘Žπ‘₯)βˆ’1.

Using a similar technique, it is also possible to derive the formula for the inverse secant function: ddsecπ‘₯ο€Ήπ‘Žπ‘₯=1π‘Ž|π‘₯|√(π‘Žπ‘₯)βˆ’1.

We can summarize our results thus far as follows.

Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions are as follows: ddsinddcosddtanddcotddsecddcscπ‘₯ο€Ίπ‘Žπ‘₯=π‘Žβˆš1βˆ’(π‘Žπ‘₯),π‘₯ο€Ήπ‘Žπ‘₯=βˆ’π‘Žβˆš1βˆ’(π‘Žπ‘₯),π‘₯ο€Ήπ‘Žπ‘₯=π‘Ž1+(π‘Žπ‘₯),π‘₯ο€Ήπ‘Žπ‘₯=βˆ’π‘Ž1+(π‘Žπ‘₯),π‘₯ο€Ήπ‘Žπ‘₯=1π‘Ž|π‘₯|√(π‘Žπ‘₯)βˆ’1,π‘₯ο€Ήπ‘Žπ‘₯=βˆ’1π‘Ž|π‘₯|√(π‘Žπ‘₯)βˆ’1.

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, ddcotddtanπ‘₯ο€Ήπ‘Žπ‘₯=βˆ’π‘₯ο€Ήπ‘Žπ‘₯.

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 ddcotπ‘₯ο€Ό1π‘₯.

Answer

Since we have a composition of functions, we will apply the chain rule. Remember, the chain rule states that, given 𝑦=𝑓(𝑒) and 𝑒=𝑔(π‘₯), dddddd𝑦π‘₯=𝑦𝑒×𝑒π‘₯.

Setting 𝑒=1π‘₯ and 𝑦=𝑒cot, we have dddd𝑦𝑒=βˆ’11+𝑒=βˆ’11+𝑒π‘₯=βˆ’1π‘₯.οŠ¨οŠ§ο—οŠ¨οŠ¨

Substituting these expressions into the formula for the chain rule, we have dd𝑦π‘₯=ο„βˆ’11+ο€»ο‡οο€Όβˆ’1π‘₯=1π‘₯+1.οŠ§ο—οŠ¨οŠ¨οŠ¨

Notice that the last example showed us that ddcotddtanπ‘₯ο€Ό1π‘₯=π‘₯π‘₯.

This is, in fact, no accident, and cottanοŠ±οŠ§οŠ±οŠ§ο€Ό1π‘₯=π‘₯.

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 ddsinπ‘₯ο€»βˆš1βˆ’π‘₯ο‡οŠ±οŠ§οŠ¨.

Answer

We have been asked to differentiate the compositions of two functions. Therefore, we should use the chain rule, which states that, given 𝑦=𝑓(𝑒) and 𝑒=𝑔(π‘₯), dddddd𝑦π‘₯=𝑦𝑒×𝑒π‘₯.

Firstly, we need to identify our 𝑒. 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 𝑒=√1βˆ’π‘₯ and 𝑦=𝑒sin. We now differentiate each of these. Beginning with 𝑦, we have dd𝑦𝑒=1√1βˆ’π‘’.

Substituting in 𝑒=√1βˆ’π‘₯, we have dd𝑦𝑒=1√1βˆ’(1βˆ’π‘₯)=1π‘₯.

We now differentiate 𝑒 with respect to π‘₯. To do this, we will need to use the chain rule again. This time we will set 𝑣=1βˆ’π‘₯ and 𝑒=βˆšπ‘£. Differentiating both 𝑒 and 𝑣, we have dddd𝑒𝑣=12βˆšπ‘£=12√1βˆ’π‘₯𝑣π‘₯=βˆ’2π‘₯.

Using the chain rule, dddddd𝑒π‘₯=𝑒𝑣×𝑣π‘₯, we have dd𝑒π‘₯=12√1βˆ’π‘₯Γ—(βˆ’2π‘₯)=βˆ’π‘₯√1βˆ’π‘₯.

We now have an expression for dd𝑒π‘₯ and dd𝑦𝑒. Hence, we can apply the chain rule to find an expression for dd𝑦π‘₯ as follows: dddddd𝑦π‘₯=𝑦𝑒×𝑒π‘₯=1π‘₯Γ—βˆ’π‘₯√1βˆ’π‘₯=βˆ’1√1βˆ’π‘₯.

Once again, we have an interesting result that ddsinddcosπ‘₯ο€»βˆš1βˆ’π‘₯=π‘₯π‘₯.

In a similar way to our last example, this result actually comes from the identity sincosοŠ±οŠ§οŠ¨οŠ±οŠ§ο€»βˆš1βˆ’π‘₯=π‘₯, for 0≀π‘₯≀1. 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: ddsinddcosddtanddcotddsecddcscπ‘₯ο€Ίπ‘Žπ‘₯=π‘Žβˆš1βˆ’(π‘Žπ‘₯),π‘₯ο€Ήπ‘Žπ‘₯=βˆ’π‘Žβˆš1βˆ’(π‘Žπ‘₯),π‘₯ο€Ήπ‘Žπ‘₯=π‘Ž1+(π‘Žπ‘₯),π‘₯ο€Ήπ‘Žπ‘₯=βˆ’π‘Ž1+(π‘Žπ‘₯),π‘₯ο€Ήπ‘Žπ‘₯=1π‘Ž|π‘₯|√(π‘Žπ‘₯)βˆ’1,π‘₯ο€Ήπ‘Žπ‘₯=βˆ’1π‘Ž|π‘₯|√(π‘Žπ‘₯)βˆ’1.
  • Familiarity with trigonometric identities can oftentimes significantly simplify the process of finding derivatives.

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