[MUSIC] What's the derivative of arc 
sine? 
Well, I can't really compute this 
directly. 
I don't know how to differentiate arc 
sine right off the bat. 
So, I'm going to try to sneak up on the 
derivative of arc sine. 
Now, instead of writing down arc sine all 
the time, I'm going to give it a 
different name. 
I'll write f of x for arc sine x. 
It just, it looks shorter. 
Now, what do I know about arc sine? 
Arc sine is the inverse function for 
sine. 
So, at least for a range of values of x, 
f of sine x is x, 
right? 
Arc sine tells me a value that I can take 
the sine of to get the input to arc sine, 
so if I evaluate arc sine at sine x, I 
get x, at least over a particular range 
of values of x. 
So if this is true, 
and if f were differentiable, I could 
apply the chain rule and differentiate 
both sides. 
So, let's just do that. 
I'm just plowing ahead assuming that f is 
differentiable. 
So, if I assume that f is differentiable, 
I differentiate f. 
I get some mystery derivative. 
I don't know what it is yet, 
I'm calling it f prime. 
At the inside, which is sine x, times the 
derivative of the inside, which is cosine 
x. 
And this is equal to the derivative of 
the other side which is one. 
So, what I know now is that the 
derivative of arc sine at sine x is one 
over cosine x. 
Try to divide both sides by cosine. 
But, what I really want is a formula that 
tells me the derivative of arc sine at 
some point is some other point, alright? 
I don't want to know the derivative at 
sine of x, in terms of cosine. 
But, I can use a trick, right? sine 
squared plus cosine squared is one. 
So if I knew that cosign were positive, I 
could rewrite this and you should be a 
little skeptical as the quality, is one 
over the square root of one minus sine 
squared x. 
You have to be a little bit worried here 
because you got to know that cosine's 
positive. 
Okay. 
But let's suppose it is. So, if cosine's 
positive, then I can write cosine x as 
the square root of one minus sine squared 
x, and now I've got a formula for the 
derivative of arc tan, derivative of f, 
at sine x is something involving sine x. 
So, I could rewrite this formula as just 
the derivative of arc sine at x, is one 
over the square root of one minus x 
squared. 
So, there we go. 
There is a formula for arc sine assuming 
that arc sine is differentiable. 
Alright? 
Because what I did is I just wrote down a 
function, you know, that I knew to be 
true. 
I know that fsinx sine x is x for certain 
values of x. 
And I just plowed through, 
differentiating it, assuming that it was 
differentiable. 
And then, I figured out what a derivative 
would have to be if it were 
differentiable. 
But note that I never actually verified 
that arc sine is differentiable. 
I'm just telling you what the derivative 
is if it were differentiable. 
So that's perhaps a little bit 
unfortunate. 
Alright. 
But nevertheless we're going to use this 
and this is in fact, the derivative of 
arc sine. 
Now, we know how to differentiate arc 
sine. 
Let's try to differentiate arc cosine. 
We could do that in the same way. 
f equals arc cosine to the chain rule 
trick. 
We get a formula for the derivative of 
arc cosine. 
Assuming it's differentiable, and it is. 
But, I want to do it a different way. 
I want to try to do it different attack, 
a different approach on the problem of 
calculating the derivative of arc cosine. 
So, here's how we're going to start. 
Take a look at this triangle here. 
this angle's alpha, and I'm labeling this 
side to have length y, 
and the hypotenuse of this right triangle 
has length one. 
Now, what this triangle's telling me is 
that the sine of alpha, which is this 
sign divided by hypotenuse, is y. 
And consequently, the arc sine of y is 
alpha. 
Now, what's arc cosine of y? 
To figure that out let's draw another one 
of these triangles. 
I'm going to label this side beta, and 
I'm going to imagine picking up this 
right triangle and flipping it over. 
So that then this angle's beta, this 
angle's alpha, hypotenuse is still length 
one, but the side that I labeled y is now 
down here. 
And what this triangle is telling me is 
that the cosine of beta, right? Which is 
this width divided by the length 
hypotenuse is is y. 
And consequently, the arc cosine of y is 
beta. 
Why is that significant? 
Well, this is a right triangle and the 
angles add up to 180 degrees, so alpha 
plus beta add up to 90 degrees, or pi 
over two radiants. 
So, that means what? 
I know that alpha plus beta is equal to 
pi over two. 
Consequently, arc sine y plus arc cosine 
y, which is alpha plus beta, add up to pi 
over two. 
Now, this formula is, you know, 
interesting in its own right. 
But think about what it means 
differentiably. 
I just calculated the derivative of arc 
sine and I'm wondering, what's the 
derivative of arc cosine? 
Well, whatever it is, right? These two 
functions add up to a constant, they're 
pi over two. 
So, however wiggling y must affect arc 
sine, wiggling y must affect arc cosine 
in precisely the opposite way so as to 
make this sum a constant. 
Said differently, right? 
The derivative of this side has to be the 
derivative of this side. 
But the derivative of this side is zero, 
that means the derivative of arc sine 
plus the derivative of arc cosine must be 
zero. 
That means that the derivative of arc 
sine has to be exactly the negative of 
the derivative of arc cosine. 
And since we know the derivative arc sine 
is one over the square root of one minus 
x squared, that means the derivative of 
arc cosine is negative one over the 
square root of one minus x squared. 
My goal right now is to differentiate arc 
tan x, but that's too hard. 
So instead, we're going to sort of sneak 
up on the derivative of arc tan x and 
figure out the derivative without 
explicitly knowing it to begin with. 
So, here we go. 
I'll make this a little bit easier, I 
don't want to keep writing arc tan x all 
the time. 
So, let's give it a different name. 
let's just call it f. 
So f of x will be arc tan x for the time 
being. 
Now, what do I know about arc tan? 
Arc tan is one of these inverse 
trigonometric functions. 
It's the inverse to tan. 
So, I know what f of tangent x is. 
Well, as long as x is between minus pi 
over four and pi over four, the arc 
tangent of tangent is just x. 
Now, this is a really great thing, right? 
Now if, if f were differentiable, I could 
use the chain rule on this. 
So, let's just pretend that arc tan is 
differentiable and see what happens. 
So, in that case, I differentiate this 
using the chain rule. 
I got the derivative of f at the inside, 
times the derivative of the inside. 
The derivative of tangent is secant 
squared. 
[SOUND] The derivative of x is one, 
right? 
So, I differentiated both sides of this 
equation, 
f tangent x equal to x. 
The derivative of f tangent x is f prime 
tangent x times the derivative of tangent 
secant squared. 
And the derivative of x, with respect to 
x, is just one. 
Now, I'll divide both sides by secant 
squared x. 
And I find out that f prime of tangent x 
is one over secant squared x. 
Now, what do I do? 
I, I'm trying to write down the 
derivative of f. 
f is arc tan, and I want to know how to 
differentiate arc tan. and what I've 
learned is that if f were differentiable, 
its derivative at tangent x would be one 
over secant squared x. 
Well, here's a trick. 
I know the Pythagorean identity. 
I know that sine squared plus cosine 
squared is one. 
So, if I take this identity and divide 
both sides by cosine squared, I get that 
tangent squared, 
right? That's sine squared over cosine 
squared, plus cosine squared over cosine 
squared one, 
is one over cosine squared, that's a 
secant squared. 
So, here's another trig identity, tangent 
squared x plus one is secant squared x. 
So, up here where I've got one over 
secant squared x, I could have written 
this as one over, instead of secant 
squared x, I'll use the fact that secant 
squared x is tangent squared x plus one. 
I'll put that in here and I get that the 
derivative of f, the derivative arc tan, 
at tangent x is one over tangent squared 
x plus one. 
Now, this is a pretty great formula, 
right? 
This is saying that the derivative at 
something is one over that same thing 
squared plus one. 
So a more normal, a more common way of 
seeing this written is that the 
derivative of arc tan x is one over x 
squared plus one. 
So, here's what we've done. 
We've got the derivative of arc sine, is 
one over the square root of one minus x 
squared. 
The derivative of arc cosine is exactly 
negative that, because the sum of arc 
sine and arc cosine is a constant, 
-1 over the squared of one minus x 
squared. 
And we've calculated the derivative of 
arc tangent just now to be one over one 
plus x squared. 
Excellent. 
At this point, we've calculated the 
derivatives of arc sine, arc cosine, and 
arc tangent. 
And now I challenge you to go and 
calculate the derivatives of arc secant, 
arc cosecant, and arc cotangent. 
Have fun. 
[MUSIC]