If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: AP®︎/College Calculus BC>Unit 3

Lesson 5: Differentiating inverse trigonometric functions

# 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).
• 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.
• , 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?`
• 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!
• At why we do not directly use the identity (tanx)^2+1 = (secx)^2 , is it wrong to use it here ?
• 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!
• 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).
• 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.
• 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).
• 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.
• Hello, why is it useful to express the derivative as a function of x?
• 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).
• 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?
• The chain rule will need to be applied. (f(g(x))' will always equal g'(x)*f'(g(x)).
• At if tan(y) = x = 1/cos^2(y) . Could not we write that cos^2(y) = (1/cos^2(y))^-1 = 1/x?