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[MUSIC] What's the derivative of arc 
sine? 

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Well, I can't really compute this 
directly. 

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I don't know how to differentiate arc 
sine right off the bat. 

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So, I'm going to try to sneak up on the 
derivative of arc sine. 

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Now, instead of writing down arc sine all 
the time, I'm going to give it a 

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different name. 
I'll write f of x for arc sine x. 

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It just, it looks shorter. 
Now, what do I know about arc sine? 

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Arc sine is the inverse function for 
sine. 

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So, at least for a range of values of x, 
f of sine x is x, 

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right? 
Arc sine tells me a value that I can take 

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the sine of to get the input to arc sine, 
so if I evaluate arc sine at sine x, I 

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get x, at least over a particular range 
of values of x. 

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So if this is true, 
and if f were differentiable, I could 

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apply the chain rule and differentiate 
both sides. 

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So, let's just do that. 
I'm just plowing ahead assuming that f is 

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differentiable. 
So, if I assume that f is differentiable, 

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I differentiate f. 
I get some mystery derivative. 

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I don't know what it is yet, 
I'm calling it f prime. 

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At the inside, which is sine x, times the 
derivative of the inside, which is cosine 

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x. 
And this is equal to the derivative of 

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the other side which is one. 
So, what I know now is that the 

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derivative of arc sine at sine x is one 
over cosine x. 

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Try to divide both sides by cosine. 
But, what I really want is a formula that 

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tells me the derivative of arc sine at 
some point is some other point, alright? 

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I don't want to know the derivative at 
sine of x, in terms of cosine. 

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But, I can use a trick, right? sine 
squared plus cosine squared is one. 

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So if I knew that cosign were positive, I 
could rewrite this and you should be a 

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little skeptical as the quality, is one 
over the square root of one minus sine 

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squared x. 
You have to be a little bit worried here 

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because you got to know that cosine's 
positive. 

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Okay. 
But let's suppose it is. So, if cosine's 

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positive, then I can write cosine x as 
the square root of one minus sine squared 

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x, and now I've got a formula for the 
derivative of arc tan, derivative of f, 

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at sine x is something involving sine x. 
So, I could rewrite this formula as just 

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the derivative of arc sine at x, is one 
over the square root of one minus x 

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squared. 
So, there we go. 

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There is a formula for arc sine assuming 
that arc sine is differentiable. 

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Alright? 
Because what I did is I just wrote down a 

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function, you know, that I knew to be 
true. 

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I know that fsinx sine x is x for certain 
values of x. 

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And I just plowed through, 
differentiating it, assuming that it was 

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differentiable. 
And then, I figured out what a derivative 

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would have to be if it were 
differentiable. 

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But note that I never actually verified 
that arc sine is differentiable. 

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I'm just telling you what the derivative 
is if it were differentiable. 

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So that's perhaps a little bit 
unfortunate. 

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Alright. 
But nevertheless we're going to use this 

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and this is in fact, the derivative of 
arc sine. 

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Now, we know how to differentiate arc 
sine. 

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Let's try to differentiate arc cosine. 
We could do that in the same way. 

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f equals arc cosine to the chain rule 
trick. 

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We get a formula for the derivative of 
arc cosine. 

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Assuming it's differentiable, and it is. 
But, I want to do it a different way. 

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I want to try to do it different attack, 
a different approach on the problem of 

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calculating the derivative of arc cosine. 
So, here's how we're going to start. 

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Take a look at this triangle here. 
this angle's alpha, and I'm labeling this 

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side to have length y, 
and the hypotenuse of this right triangle 

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has length one. 
Now, what this triangle's telling me is 

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that the sine of alpha, which is this 
sign divided by hypotenuse, is y. 

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And consequently, the arc sine of y is 
alpha. 

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Now, what's arc cosine of y? 
To figure that out let's draw another one 

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of these triangles. 
I'm going to label this side beta, and 

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I'm going to imagine picking up this 
right triangle and flipping it over. 

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So that then this angle's beta, this 
angle's alpha, hypotenuse is still length 

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one, but the side that I labeled y is now 
down here. 

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And what this triangle is telling me is 
that the cosine of beta, right? Which is 

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this width divided by the length 
hypotenuse is is y. 

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And consequently, the arc cosine of y is 
beta. 

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Why is that significant? 
Well, this is a right triangle and the 

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angles add up to 180 degrees, so alpha 
plus beta add up to 90 degrees, or pi 

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over two radiants. 
So, that means what? 

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I know that alpha plus beta is equal to 
pi over two. 

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Consequently, arc sine y plus arc cosine 
y, which is alpha plus beta, add up to pi 

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over two. 
Now, this formula is, you know, 

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interesting in its own right. 
But think about what it means 

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differentiably. 
I just calculated the derivative of arc 

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sine and I'm wondering, what's the 
derivative of arc cosine? 

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Well, whatever it is, right? These two 
functions add up to a constant, they're 

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pi over two. 
So, however wiggling y must affect arc 

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sine, wiggling y must affect arc cosine 
in precisely the opposite way so as to 

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make this sum a constant. 
Said differently, right? 

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The derivative of this side has to be the 
derivative of this side. 

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But the derivative of this side is zero, 
that means the derivative of arc sine 

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plus the derivative of arc cosine must be 
zero. 

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That means that the derivative of arc 
sine has to be exactly the negative of 

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the derivative of arc cosine. 
And since we know the derivative arc sine 

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is one over the square root of one minus 
x squared, that means the derivative of 

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arc cosine is negative one over the 
square root of one minus x squared. 

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My goal right now is to differentiate arc 
tan x, but that's too hard. 

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So instead, we're going to sort of sneak 
up on the derivative of arc tan x and 

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figure out the derivative without 
explicitly knowing it to begin with. 

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So, here we go. 
I'll make this a little bit easier, I 

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don't want to keep writing arc tan x all 
the time. 

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So, let's give it a different name. 
let's just call it f. 

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So f of x will be arc tan x for the time 
being. 

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Now, what do I know about arc tan? 
Arc tan is one of these inverse 

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trigonometric functions. 
It's the inverse to tan. 

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So, I know what f of tangent x is. 
Well, as long as x is between minus pi 

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over four and pi over four, the arc 
tangent of tangent is just x. 

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Now, this is a really great thing, right? 
Now if, if f were differentiable, I could 

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use the chain rule on this. 
So, let's just pretend that arc tan is 

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differentiable and see what happens. 
So, in that case, I differentiate this 

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using the chain rule. 
I got the derivative of f at the inside, 

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times the derivative of the inside. 
The derivative of tangent is secant 

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squared. 
[SOUND] The derivative of x is one, 

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right? 
So, I differentiated both sides of this 

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equation, 
f tangent x equal to x. 

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The derivative of f tangent x is f prime 
tangent x times the derivative of tangent 

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secant squared. 
And the derivative of x, with respect to 

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x, is just one. 
Now, I'll divide both sides by secant 

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squared x. 
And I find out that f prime of tangent x 

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is one over secant squared x. 
Now, what do I do? 

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I, I'm trying to write down the 
derivative of f. 

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f is arc tan, and I want to know how to 
differentiate arc tan. and what I've 

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learned is that if f were differentiable, 
its derivative at tangent x would be one 

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over secant squared x. 
Well, here's a trick. 

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I know the Pythagorean identity. 
I know that sine squared plus cosine 

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squared is one. 
So, if I take this identity and divide 

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both sides by cosine squared, I get that 
tangent squared, 

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right? That's sine squared over cosine 
squared, plus cosine squared over cosine 

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squared one, 
is one over cosine squared, that's a 

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secant squared. 
So, here's another trig identity, tangent 

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squared x plus one is secant squared x. 
So, up here where I've got one over 

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secant squared x, I could have written 
this as one over, instead of secant 

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squared x, I'll use the fact that secant 
squared x is tangent squared x plus one. 

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I'll put that in here and I get that the 
derivative of f, the derivative arc tan, 

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at tangent x is one over tangent squared 
x plus one. 

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Now, this is a pretty great formula, 
right? 

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This is saying that the derivative at 
something is one over that same thing 

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squared plus one. 
So a more normal, a more common way of 

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seeing this written is that the 
derivative of arc tan x is one over x 

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squared plus one. 
So, here's what we've done. 

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We've got the derivative of arc sine, is 
one over the square root of one minus x 

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squared. 
The derivative of arc cosine is exactly 

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negative that, because the sum of arc 
sine and arc cosine is a constant, 

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-1 over the squared of one minus x 
squared. 

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And we've calculated the derivative of 
arc tangent just now to be one over one 

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plus x squared. 
Excellent. 

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At this point, we've calculated the 
derivatives of arc sine, arc cosine, and 

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arc tangent. 
And now I challenge you to go and 

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calculate the derivatives of arc secant, 
arc cosecant, and arc cotangent. 

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Have fun. 
[MUSIC]