Video: Derivatives of Inverse Trigonometric Functions

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

15:56

Video Transcript

In this video, we’ll learn how to find the derivatives of the inverses of trigonometric functions. We will learn how to do this by using implicit differentiation. And thus, it is important that you understand how to apply the chain rule, if not have a thorough understanding of how implicit differentiation works, before watching this video. After deriving the derivatives of the inverse trigonometric functions, we’ll consider the application of these derivatives to more complicated inverse trigonometric functions.

Before we look at finding the derivative of our inverse trigonometric functions, let’s just quickly consider the function 𝑦 equal sin of π‘₯. Remember, π‘₯ is a real number. And of course, since we’re performing calculus with a trigonometric function, we need to make sure this is measured in radians. For this function, we can say that π‘₯ is equal to the inverse sin of 𝑦. This subscript negative one denotes the inverse function. Remember though without restricting the domain of arc sin of π‘₯ or inverse sin of π‘₯, the function is going to be many to one. We, therefore, restrict the domain for 𝑓 of π‘₯ equals inverse sin of π‘₯. And we say that π‘₯ has to be greater than or equal to negative one and less than or equal to one.

If we go back to the function π‘₯ equals the inverse sin of 𝑦, we can, therefore, see that π‘₯ will take values greater than or equal to negative πœ‹ by two and less than or equal to πœ‹ by two. We’ll also need to recall the chain rule. This says that if 𝑦 is some function in 𝑒 and 𝑒 is some differentiable function in π‘₯, then d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. We’ll now use everything we’ve seen here to find the derivative of the inverse sine function.

Find the derivative of the inverse sine function with respect to π‘₯.

We’re differentiating the inverse sin of π‘₯ with respect to π‘₯. So we’re going to begin by letting 𝑦 be equal to the inverse sin of π‘₯. Then we can say that π‘₯ must be equal to sin of 𝑦. We’re going to differentiate both sides of this equation with respect to π‘₯. So we say that d by dπ‘₯ of π‘₯ is equal to d by dπ‘₯ of sin of 𝑦. Well, the derivative of π‘₯ with respect to π‘₯ is quite straight forward; it’s one. But we’re going to need to use implicit differentiation which is a special case of the chain rule to differentiate sin 𝑦 with respect to π‘₯.

The derivative of sin 𝑦 with respect to 𝑦 is cos 𝑦. So the derivative of sin 𝑦 with respect to π‘₯ is cos 𝑦 times the derivative of 𝑦 with respect to π‘₯ which is just d𝑦 by dπ‘₯. So we currently see that one is equal to cos of 𝑦 times d𝑦 by dπ‘₯. We divide both sides of this equation by cos 𝑦 to form an equation for the derivative. And we see that d𝑦 by dπ‘₯ is equal to one over cos 𝑦. Now, we’ve got a bit of a problem. We do want an expression for the derivative in terms of π‘₯ not 𝑦.

And remember, we said that π‘₯ was equal to sin of 𝑦. So we’ll use the identity cos squared πœƒ plus sin squared πœƒ equals one. And I’ve replaced πœƒ with 𝑦. We’ll subtract sin squared 𝑦 from both sides of the equation. And then we’ll take the square root of both sides. And we see that cos of 𝑦 is equal to the positive and negative square root of one minus sin squared 𝑦. Remember, inverse sin is restricted to the closed interval negative πœ‹ by two to πœ‹ by two.

By our definition, that means 𝑦 must be greater than or equal to πœ‹ two and less than or equal to πœ‹ by two which, in turn, means cos of 𝑦 must be greater than or equal to zero and less than or equal to one. And that’s because in the interval 𝑦 is greater than or equal to negative πœ‹ by two and less than or equal to πœ‹ by two. The smallest value cos of 𝑦 we’ll take is zero. And the greatest value is one. And what this means here is we’re going to take the positive square root of one minus sin squared 𝑦 only.

We can now replace sin of 𝑦 with π‘₯. And we see that cos of 𝑦 is equal to the square root of one minus π‘₯ squared. And, therefore, d𝑦 by dπ‘₯ equals one over the square root of one minus π‘₯ squared. And we found the derivative of the inverse sin of π‘₯. It’s one over the square root of one minus π‘₯ squared for values of π‘₯ in the range π‘₯ is greater than negative one and less than one. In our next example, we’ll consider an alternative method that will help us find the derivative of the inverse cosine function.

This time, we’re going to need to know the inverse function theorem. This says that if 𝑓 is a differentiable function with a continuous inverse 𝑓 prime and 𝑓 prime of π‘Ž is not equal zero, then not only is 𝑓 invertible but it has a differentiable inverse. Such that the derivative of the inverse of 𝑓 at some 𝑏 equals 𝑓 of π‘Ž is equal to one over the derivative of 𝑓 at π‘Ž. This is sometimes written simply as dπ‘₯ by d𝑦 is equal to one over d𝑦 by dπ‘₯. Let’s see how this might help us when differentiating the inverse cosine function.

Find the derivative of the inverse cos of π‘₯ over π‘Ž with respect to π‘₯, where π‘Ž is not equal to zero.

We’ll begin by letting 𝑦 be equal to the inverse cos of π‘₯ over π‘Ž. This can be alternately written as π‘₯ over π‘Ž equals cos of 𝑦. And we can then multiply both sides by π‘Ž. And we see that π‘₯ is equal to π‘Ž times cos of 𝑦. We’re going to differentiate our expression for π‘₯ with respect to 𝑦. In other words, we’re going to find dπ‘₯ by d𝑦. We’ll use the general result that the derivative of cos of π‘₯ with respect to π‘₯ is negative sin of π‘₯. And we see that dπ‘₯ by d𝑦 must be equal to negative π‘Ž sin of 𝑦.

Now, before we perform the next step we need to recall the fact that for the inverse trigonometric functions, we restrict their domains. And we know that the domain of the inverse cos of π‘₯ or our cos of π‘₯ is greater than or equal to zero and less than or equal to πœ‹. This means that 𝑦 must be greater than or equal to zero and less than or equal to πœ‹. Now, we’re going to use the inverse function theorem. So we’re going to take values of 𝑦 greater than zero and less than πœ‹, such that sin of 𝑦 is not equal to zero.

Using this criteria, we can use dπ‘₯ by d𝑦 equals one over d𝑦 by dπ‘₯ which can be rearranged to say that d𝑦 by dπ‘₯ equals one over dπ‘₯ by d𝑦. And we see that, for our case, d𝑦 by dπ‘₯ equals one over negative π‘Ž sin of 𝑦. And we have an expression for the derivative in terms of 𝑦. Remember, we want this to be in terms of π‘₯. We said that π‘₯ over π‘Ž equals cos of 𝑦. So we’ll use the fact that sin squared 𝑦 plus cos squared 𝑦 equals one and rearrange this to say that sin 𝑦 equals plus or minus the square root of one minus cos squared 𝑦. When 𝑦 is between zero and πœ‹, sin 𝑦 is greater than zero. So in fact, we’re only interested in the positive root.

So we’ll replace this in our expression for the derivative. And we get negative one over π‘Ž times the square root of one minus cos squared 𝑦. We then replace cos 𝑦 with π‘₯ over π‘Ž and change π‘₯ over π‘Ž all squared to π‘₯ squared over π‘Ž squared. And then we bring π‘Ž into the square root. And we see that d𝑦 by dπ‘₯ is equal to negative one over the square root of π‘Ž squared minus π‘₯ squared. So d by dπ‘₯ of the inverse cos of π‘₯ over π‘Ž is equal to negative one over the square root of π‘Ž squared minus π‘₯ squared for values of π‘₯ between negative π‘Ž and π‘Ž. In our next example, we’ll consider how we might apply the process that’s used so far to find the derivative of the inverse tangent function.

Find an expression for the derivative of 𝑦 equals the inverse tan of π‘Žπ‘₯ in terms of π‘₯.

Since 𝑦 is equal to the inverse tan of π‘Žπ‘₯, we can write π‘Žπ‘₯ as being equal to tan 𝑦. We’re going to use implicit differentiation to find the derivative of both sides of this equation. The derivative of π‘Žπ‘₯ with respect to π‘₯ is simply π‘Ž. And the derivative of tan 𝑦 with respect to π‘₯ is equal to the derivative of tan 𝑦 with respect to 𝑦 times the derivative of 𝑦 with respect to π‘₯. The derivative of tan π‘₯ is sec squared π‘₯. And the derivative of 𝑦 with respect to π‘₯ is d𝑦 by dπ‘₯.

So we see that π‘Ž is equal to sec squared 𝑦 times d𝑦 by dπ‘₯. Dividing through by sec squared 𝑦 and we see that d𝑦 by dπ‘₯ equals π‘Ž over sec squared 𝑦. We’re going to need to represent our equation for the derivative in terms of π‘₯. So we’ll use this trigonometric identity. One plus tan squared π‘₯ equals sec squared π‘₯. This means we can write d𝑦 by dπ‘₯ as π‘Ž over one plus tan squared 𝑦. And then we replaced tan 𝑦 with π‘Žπ‘₯. And we see the expression for the derivative of 𝑦 equals the inverse tan of π‘Žπ‘₯ is π‘Ž over one plus π‘Žπ‘₯ squared.

Similar rules can be applied to help us find the derivative of the inverse cotangent function. We find that the derivative of the inverse of cotangent of π‘Ž of π‘₯ is equal to negative π‘Ž over one plus π‘Žπ‘₯ squared. The inverse cosecant and secant functions are a little more unusual. So we’ll consider next how to find the derivative of the inverse cosecant function. Find d by dπ‘₯ of the inverse cosecant of π‘₯. We begin by letting 𝑦 be equal to the inverse cosecant of π‘₯. And this means we can rewrite this. And we can say that π‘₯ is equal to the cosecant of 𝑦.

We’re next going to use implicit differentiation to find the derivative of both sides of this equation. The derivative of π‘₯ with respect to π‘₯ is simply one. Then the derivative of cosec 𝑦 with respect to π‘₯ is equal to the derivative of cosec 𝑦 with respect to 𝑦 times d𝑦 by dπ‘₯. And the derivative of cosec 𝑦 with respect to 𝑦 is negative cosec 𝑦 cot 𝑦. So we see that one is equal to negative cosec 𝑦 cot 𝑦 times d𝑦 by dπ‘₯.

Now, we know that, for the inverse cosecant function, 𝑦 must be greater than negative πœ‹ by two and less than πœ‹ by two and not equal to zero. Using these restrictions cosec 𝑦 cot 𝑦 cannot be equal to zero. So we can divide through by negative cosec 𝑦 cot 𝑦. And we see that d𝑦 by dπ‘₯ is as shown. We want to represent our equation for the derivative in terms of π‘₯. So we’ll use this trigonometric identity cot squared 𝑦 plus one equals cosec squared 𝑦. And we can rewrite this to say that cot of 𝑦 is equal to the positive and negative square root of cosec squared 𝑦 minus one.

We put this into the equation for the derivative in place of cot of 𝑦. And we then use the fact that π‘₯ is equal to cosec 𝑦. But we are going to need to make a decision on the sine of the derivative. And it can help here to look at the graph of the inverse cosecant function. Notice how, for all values of π‘₯ in the range of the function, the derivative of the slope of the tangent is negative.

And we, therefore, use the absolute value to ensure that our derivative is always negative. We say that d𝑦 by dπ‘₯ is equal to the negative of the absolute value of one over π‘₯ times the square root of π‘₯ squared minus one. Since one and the square root of π‘₯ squared minus one are always positive, we can rewrite this as shown. So the derivative of the inverse cosecant function π‘₯ with respect to π‘₯ is negative one over the modulus or absolute value of π‘₯ times the square root of π‘₯ squared minus one.

A similar process can be applied to help us find the derivative of the inverse secant function. And we have the derivatives of all of the inverse trigonometric functions we require. It’s useful to commit these results to memory but also be prepared to derive them where necessary. We’ll now have a look at the application of these results.

Evaluate the derivative of the inverse cotangent of one over π‘₯ with respect to π‘₯.

Here, we have a function of a function or a composite function. We’re, therefore, going to need to use the chain rule to find the derivative. This says that if 𝑓 and 𝑔 are differentiable functions such that 𝑦 is 𝑓 of 𝑒 and 𝑒 is 𝑔 of π‘₯, then d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. We’ll let 𝑒 be equal to one over π‘₯. Then 𝑦 is equal to the inverse cot of 𝑒. To apply the chain rule, we need to find the derivative of both of these functions. And with 𝑒 it can be useful to write it as π‘₯ to the negative one.

Then d𝑒 by dπ‘₯ is negative π‘₯ to negative two or negative one over π‘₯ squared. We can then use the general derivative of the inverse cotangent function. And we see that d𝑦 by d𝑒 is equal to negative one over one plus 𝑒 squared. d𝑦 by dπ‘₯ is the product of these. It’s negative one over π‘₯ squared times negative one over one plus 𝑒 squared.

We can replace 𝑒 with one over π‘₯ and then multiply through. And we see that the derivative of the inverse cotangent of one over π‘₯ with respect to π‘₯ is one over π‘₯ squared plus one. Did you notice that the derivative of the inverse cotangent of one over π‘₯ is equal to the derivative of the tangent of π‘₯? This is, in fact, no accident. And we can use the identity the inverse cotangent of one over π‘₯ equals the inverse tan of π‘₯. That could have saved us a little bit more time in this previous example.

Evaluate the derivative of the inverse sin of the square root of one minus π‘₯ squared with respect to π‘₯.

Here, we have a function of a function or a composite function. So we’ll use the chain rule to find its derivative. This says that if 𝑦 is some function in 𝑒 and 𝑒 is some function in π‘₯, then d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. We’ll let 𝑒 be equal to the square root of one minus π‘₯ squared. Which can, of course, alternatively be written as one minus π‘₯ squared to the power of one-half. Then 𝑦 is equal to the inverse sin of 𝑒. To apply the chain rule, we’re going to need to find the derivative of both of these functions. The derivative of the inverse sin of 𝑒 with respect to 𝑒 is one over the square root of one minus 𝑒 squared.

And we can use the general power rule to find the derivative of one minus π‘₯ squared to the power of one-half. It’s a half times one minus π‘₯ squared to the negative one-half times the derivative of the bit inside the brackets which is negative two π‘₯. That can be written as negative π‘₯ times one minus π‘₯ squared to the power of negative one-half.

d𝑦 by dπ‘₯ is, therefore, negative π‘₯ over the square root of one minus π‘₯ squared times one over the square root of one minus 𝑒 squared. We can replace 𝑒 with one minus π‘₯ squared to the power of one-half. And the second fraction becomes one over the square root of one minus one minus π‘₯ squared. This further simplifies to one over π‘₯. And we divide through by π‘₯. And we see that the derivative of our function is negative one over the square root of one minus π‘₯ squared.

Once again, we’ve stumbled across an interesting result. That is that the derivative of the inverse sin of the square root of one minus π‘₯ squared is equal to the derivative of the inverse cos of π‘₯. This comes from the identity of the inverse sin of the square root of one minus π‘₯ squared is equal to the inverse cos of π‘₯, the values of π‘₯ between zero and one. Being familiar with this result could have reduced the amount of what we needed to do in this example.

In this video, we’ve seen that we can use implicit differentiation or the inverse function theorem to derive the formulae for the derivatives of inverse trigonometric functions. We saw that the derivatives of the inverse trigonometric functions are as shown. And we also saw that being familiar with certain trigonometric identities can sometimes significantly reduce the process of finding these derivatives.

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