[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]