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[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.