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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.

AndÂˇÂ 1 0355. is awfully close to 1.0349. So we've

successfully used the derivative to approximate the value of tangent 46

degrees.