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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 32: Inverse trig functions differentiation

# Derivative of inverse tangent

Now let's explore the derivative of the inverse tangent function. Starting with the derivative of tangent, we use the chain rule and trigonometric identities to find the derivative of its inverse. Join us as we investigate this fascinating mathematical process! Created by Sal Khan.

## Want to join the conversation?

• Why not just leave the secant as it is? Using the identity relating square secant and square tangent, you can reduce 1/sec^2 y to 1/(tan^2 y +1) which is equal to 1/(x^2 + 1).
(62 votes)
• When you say "leave the secant as it is" I assume you mean when we've determined that dy/dx = cos² y, you would replace cos²y with 1/sec²y. I would agree that this produces a shorter route to the end result, and I'm not sure whether Sal overlooked this possibility or chose the longer method to demonstrate some additional solving methods.

Usually the easiest way to handle these when you get to this stage is to draw a right triangle, labeling one of the acute angles y and two of the sides in a way that corresponds to the main equality we're working with. In this case it's tan y = x, so we'd label the opposite side x and the adjacent side 1. Then it's easy to see that cos y is 1/sqrt(1 + x²). Square that and you're done.
(53 votes)
• At why we do not directly use the identity (tanx)^2+1 = (secx)^2 , is it wrong to use it here ?
(26 votes)
• Sal really likes showing the derivation of things, but yes, you could have used the identity
tan²x + 1 = sec²x and gotten the same result. Good eye!
(21 votes)
• , is it a bit convenient to divide the numerator and the denominator by `cos y`? What if `cos y = 0`? Can someone make it clear for me?`
(20 votes)
• You're right that normally you want to be careful not to divide by something which could take on a value of zero! Let's see why cos(y) can't be zero here.
Consider that Sal defined y as `y=arctan(x)`, which we know has a range of `(-pi/2, pi/2)`. But if we think about taking the cosine of values in this range, we see that cos(y) can only take on positive values. Therefore cos(y) cannot equal 0. (Note: it would also be bad since after differentiating tan(y) with respect to x, Sal obtains an expression that contains 1/(cos(y))^2. So we wouldn't want cos(y) to be zero in that differentiated equation either!) Hope that helps!
(31 votes)
• At , Sal wants to write dy/dx in terms of x, so why doesn't he just subsitute arctan(x) for the y. After all, y does equal arctan(x).
(6 votes)
• Sal wants to show why the derivative of arctan(x) is 1/(1+x^2), and this method is the easiest way of doing so. Although there probably is a way to simplify cos^2(arctan(x)) to 1/(1+x^2) , I think Sal's way was simplest.

And if you are wondering why it is so important to simplify it to the exact form of 1/(1+x^2), it is because that is the most commonly used form in textbooks, classrooms, and life.
(5 votes)
• When we get to dy/dx=(cos y)^2, is this approach viable:
Since tan y=x, the tan ratio opposite/adjacent tells you that your opposite side is x and adjacent side is 1. Now use pythagorean theorem to find the hypoteneuse, which is sqrt(x^2+1). Then form cos y= 1/sqrt(x^2+1) and sub. it back into the above formula, squaring it to give you 1/(1+x^2).
(6 votes)
• That is a great approach. It should be noted that there are some technicalities that are glossed over, but none of them affect the final result.
(3 votes)
• Hello, why is it useful to express the derivative as a function of x?
(2 votes)
• Remember that we are treating y as the dependent variable. You input a value of x, and you get a value of y. That is, y is "a function of x." When you express a derivative "with respect to x," as in dy/dx, you are asking the question, "what is the slope of the line tangent to the y value for a given value of x." In order to answer that question explicitly, you need the derivative to be expressed as a function of x so that you can "input" a value of x and calculate the derivative of y (the slope of the line tangent to y at a given value of x).
(4 votes)
• Around , wouldn't it just be easier to leave 1/cos^2(y) as sec^2(y)

then go to the other Pythagorean identity (Pythagorean corollary, if you will) that

*tan^2(y)+1 = sec^2(y)* (divide every term by cos^2(y)),

then just sub X in for tan(y) (because x = tan(y))

getting to 1/(1+x^2) in considerably less time?
(3 votes)
• Yeah, you could do that. That's basically functionally equivalent to what Sal did, but in kind of a different order.
(2 votes)
• Does the chain rule apply here? if I have the derivative of the inverse tangent of x^2 will I need to multiply by a 2x a the end?
(2 votes)
• The chain rule will need to be applied. (f(g(x))' will always equal g'(x)*f'(g(x)).
(2 votes)
• At if tan(y) = x = 1/cos^2(y) . Could not we write that cos^2(y) = (1/cos^2(y))^-1 = 1/x?
(2 votes)
• Don't the two cos^2y's cancel out in the denominator? How is Sal then dividing the sin^2y by cos^2y? I'am not understanding that process.
(0 votes)
• No, the cos²y in the denominator is being added to sin² y, so you cannot cancel out the other cos² y because you can only cancel factors, not terms.
(4 votes)

## Video transcript

We already know that the derivative with respect to x of tangent of x is equal to the secant of x squared, which is of course the same thing of one over cosine of x squared. Now what we wanna do in this video, like we've done in the last few videos, is figure out what the derivative of the inverse function of the tangent of x is, or in particular, let's see if we can figure out what the derivative with respect to x of the inverse tangent of x is. And I encourage you to pause this video and use a technique similar to the one, or very close to the one that we've used in the last two videos, to figure out what this is. Well let's set y equal to the inverse tangent of x, y is equal to inverse tangent of x. That is the same thing as saying that the tangent of y, the tangent of y is equal to x. All I've done, now you can kind of think of it as I've just taken the tangent of both sides right over here, and now we can take the derivative of both sides with respect, the derivative of both sides with respect to x. And on the left hand side, we can just apply the chain rule. Derivative of tangent of y with respect to y, is going to be secant squared of y, which is the same thing as one over cosine of y squared. I like to write it this way. It's it'll, keeps it a little bit simpler at least in, in my brain, but when we're applying the [INAUDIBLE], it's gonna be the derivative of tangent of y with respect to y times the derivative of y with respect to x, times the derivative of y with respect to x, and on the right hand side, the derivative of x with respect to x, well that's just going to be equal to one. And so we get, if we wanna solve for the derivative y with respect to x, we just multiply both sides times the cosine of y squared. And we get the derivative of y with respect to x is equal to cosine of y squared. And like we've seen in previous videos, this isn't that satisfying because, you know, I've written the derivative of y with respect to x as a, as a function of y. But what we're really interested in is writing it as a function of x, and to do that, we'd express this somehow in terms of the tangent of y. And the reason why the tangent of y is interesting is because we already know that tangent of y is equal to x. So we can rewrite this using a little bit of trigonometric identities. Then we can, with tangent of y, we can substitute all the tangent of y's with an x. So let's see if we can do that. And this seems a little bit tricky. They introduce a tangent of y, we'd wanna have a sine, divided by a cosine, that's what tangent is, and this is just a straight up cosine squared y. So this is really gonna take a little bit more experimentation than at least some of the other, the, than the last two examples we've done. So one thing we could do is we could say hey, let's just divide by one. Dividing by one never hurt anyone. So we could say this is the same thing, this is the same thing as cosine squared y. And I'm really doing this to, to see if I can in, start to express it as some type of a rational, a rational expression which might involve, at some point, a sine divided by cosine and it could have a tangent. So, let's divide by one. But we know from the Pythagorean, the Pythagorean identity that one is equal to sine squared of y plus cosine squared of y, and so let's try that or we could write cosine squared of y plus sine squared of y. Once again, why was I able to divide by this expression? Well, this expression by the Pythagorean identity, which really comes out of the unit circle definition of trig functions, this is equal to one, so I have not changed the value of this expression. Now what makes this interesting is if I wanted to introduce a sine divided by a cosine, I could just divide the numerator and the denominator by cosine square, so lets do that, lets multiply one over cosine or else divide the numerator by cosine square of y and divide the denominator, by cosine square of y, or multiply each of them by one over cosine square of y. What's that going to give us? Well the numerator, these characters are going to cancel you're just going to have a one. And the denominator this time this this, that's going to be equal to one. And then you're gonna have sine squared, sine squared y over, over cosine squared y, and this is, this is the goal that I was trying to achieve, I have a sine divided by cosine squared. So this right over here, this is the same thing, actually let me just write it this way, let me, let me write it this way. This is the same thing as sine of y over cosine of y, cosine of y. Whole thing squared, which is of course the same thing as one over one plus tangent of y, tangent of y squared. This is equal to this. Now why is that useful? Well we know that x is equal to tangent of y. So this is going to be equal to, this is going to be equal to one, one over, one plus tangent of y is equal to x, x squared, x squared, which is pretty exciting. We just figured out the derivative of y with respect to x. So the derivative of this thing with respect to x is one over one plus x squared. So we could write that right up here. So this is going to be equal to one over one plus x squared, and we are done.