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## Differentiating inverse trigonometric functions

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# Derivative of inverse tangent

AP Calc: FUN‑3 (EU), FUN‑3.E (LO), FUN‑3.E.2 (EK)

## 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.