I want to differentiate tangent theta. 
How am I going to do this? 
Well, hopefully you remember that tangent 
theta is sine theta over cosine theta. 
Why is this? 
If you think back to the business about 
the right triangles, right, sine is this 
opposite side that I'm calling y over the 
hypotenuse. 
And cosine is this adjacent side whose 
length I'll call x over the hypotenuse. 
I'll call this angle theta. 
Then sine theta is y over r, and cosine 
theta is x over r, so this fraction can 
be simplified to y over x. 
That's the opposite side over the 
adjacent side. 
That's tangent of theta. 
So good. 
I can express Tangent theta as sin theta 
over cosine theta, how does that help? 
Well, I know how to differentiate sine 
and cosine, and by writing tangent this 
way, tangents now a quotient, so I can 
use the. 
Quotient rule. 
All right, so I'll use the quotient rule, 
and the derivative of tangent theta is 
the derivative of the numerator, which is 
cosine, times the der, denominator, which 
is just cosine, minus the derivative of 
the denominator, which is minus sine, 
times the numerator, which is sine. 
And this whole thing is over the 
denominator squared, cosine squared 
theta. 
Now, is this very helpful? 
Well, I've got cosine times cosine, so I 
can write that as cosine squared theta. 
And here I've got minus negative sine 
theta times sine theta, so that ends up 
being plus sine squared theta. 
And the whole thing is still being 
divided by cosine squared theta. 
Can I simplify that at all? 
Well cosine squared plus sine squared, 
that's the Pythagorean identity. 
That's just one. 
So I can replace the numerator with just 
one, still over cosine squared theta. 
And one over cosine squared theta, well, 
one over cosine is called secant, so this 
is really secant squared theta. 
. 
So what all this shows is the derivative 
of tangent theta is second squared theta. 
A moment ago we did a calculation using 
the quotient rule to see that the 
derivative of tangent theta is second 
squared theta. 
And now I want to see how this plays out 
in some concrete example to get a, a real 
sense as to why a formula like this is 
true. 
So here's a couple triangles. 
They're both right triangles and the 
length of their hypotenuse is the same. 
This angle I am calling alpha, let's this 
be a little bit less than 45 degree. 
And this angle I am calling beta, and 
it's more than 45 degrees. 
Now what you can say about the Secant of 
alpha and the Secant of beta? 
You know, the Secant of alpha is 
definitely bigger than one. 
I mean, what's the Secant? 
It's this hypotenuse length divided by 
this length here. 
And that's bigger than one. 
How does it compare to beta? 
Well, the secant of beta is quite a bit 
bigger than the secant of alpha, and why 
is that? 
Well, the secant of beta is this length 
here, the hypotenuse, divided by this 
width, but this triangle is quite a bit 
narrower than this triangle. 
So the secant has the same numerator, the 
hypotenuse, the same length, but the 
denominator here is quite a bit smaller, 
and if the denominator's a lot smaller, 
then the ratio, which is the secant, is 
quite a bit bigger. 
Some of the secan of beta is bigger than 
the secan of alpha, and the secan of 
alpha is bigger than one. 
And that means that secan squared alpha 
is also bigger than one. 
And secan squared alpha is less than 
secan squared beta. 
And the significance of that is right 
here, the derivative of tangent theta is 
secan squared theta. 
So what does this mean? 
Well, this, this is telling me how 
wiggling theta affects tangent. 
It affects it like a factor of secant 
squared theta. 
So in this example, where secant squared 
beta is a lot bigger than secant squared 
alpha, the effect of wiggling beta on the 
tangent of beta should be a lot larger 
than the effect of wiggling alpha on the 
tangent of alpha. 
And you can see that. 
Let me draw a triangle, where I've 
wiggled the angle alpha up a little bit. 
I've made it a little bit bigger. 
But I'm going to make the same 
hypotenuse. 
All right, so this hypotenuse length is 
the same as this length, but I've made 
the angle a little bit bigger. 
And how is the tangent of the slightly 
larger alpha compared to the tangent of 
alpha. 
Well, the tangent of the slightly larger 
alpha is bigger than tangent alpha, but 
not by all that much. 
Now compare that to when I wiggle beta up 
by the same amount. 
I make beta a little bit bigger. 
Right. 
So I make beta a little bit bigger by the 
same amount that I made alpha larger. 
And I think about how that affects the 
tangent of beta. 
The tangent of beta is this height 
divided by, divided by this width. 
And you can think about it, I mean this, 
this height maybe isn't increasing a 
whole lot. 
But the width of this triangle is getting 
quite a bit narrower. 
And because it's the ratio of that height 
to that width the tangent of the 
perturbed value of beta is quite a bit 
larger than the tangent of beta. 
And you can, you know it's reflected in 
the fact the secant squared beta is a lot 
larger than secant squared alpha. 
So these, these kind of facts, right? 
The fact that the derivative of tangent 
is secant squared theta you, you can 
really get a sense for why these things 
might be true by thinking about triangles 
and how wiggling the angle will affect 
certain ratios of sides of the triangles. 
But, if this seems a little too abstract 
we can kind of pull back a little bit and 
do do a numeric example next. 
You know, and maybe the numerical example 
is sort of another way to see a formula 
like this in action. 
Let's do a numerical example to get a 
sense as to what you might do with the 
fact that the derivative of tangent is 
secen squared. 
Here's an example, let's try to 
approximate the value of tangent of 46 
degrees. 
Why is this an interesting example? 
Well, we know the tangent of 45 degrees 
exactly, all right. 
And figure that out by looking at a 
triangle, here's the angle, 45 degrees, a 
right triangle, because the an, angles 
add up to, 180 degrees. 
So it's 45 plus 90 plus what? 
Well, this must also be 45. 
It's an isosceles triangle now, so these 
two sides are the same. 
A tangent is the ratio of this side to 
this side because they are equal that 
ratio was one. 
So I know the tangent of 45 exactly. 
It's one. 
But I am trying to figure out an 
approximation for the tangent of 46 
degrees, the derivative tells me how 
wrigly an input affects the output, so I 
can use this fact and the fact that I 
know the derivative to try to approximate 
the tangent of 46. 
In particular the tangent of 46 degrees 
is the tangent of 45 plus one degree. 
And here you can see how I am perturbing 
the input a bit. 
And this is exactly what the derivative 
would tell me something about. 
A little bit of bad news, the derivative 
of tangent is only secant squared if I do 
the measurement in radians. 
If I convert this to a problem in 
radians. 
With radians, says the tangent of pi/4, 
which is 45 degrees, plus with one degree 
in radians, is pi/180 radians, right? 
This is what I want to compute. 
I want to compute the tangent of pi/4 
plus pi/180. 
And I'll do that with approximation using 
the derivative. 
So according to the derivative this is 
about the tangent of pi over four, which 
is the tangent of 45 degrees, it's one. 
Plus the derivative of pi over four, 
which is secant squared pi over four. 
Times how much I wiggled the input by, 
which is pi over 180. 
I know the tangent of pi over four. 
It's one. 
What's the secant of pi over four 
squared? 
Well, if I pretend that these sides have 
length one, by the pythagorean theorem, 
this side must have length square root of 
two. 
The secant is hypotenuse over this width. 
So the secant of pi over four is square 
root of two. 
So secant squared pi over four is square 
root of two squared, which is two times 
pi over 180. 
Now if you know an approximation for pi, 
you can compute two times pi over 180 
plus one. 
And this is approximately 1.0349, 
and it keeps going Pi's irrational. 
And this is not so far off of the actual 
value. 
If we actually compute tangent of 46, the 
actual value is about 1.0355 and it keeps 
going. 
And· 1 0355. 
is awfully close to 1.0349. So we've 
successfully used the derivative to 
approximate the value of tangent 46 
degrees. 
