Trig Derivatives
On this page we find , and .
We use the unit-circle definitions of and :
Interpret as an angle, and rotate that amount counter-clockwise
on the unit circle starting at .
Then is the -coordinate of the resulting point and is the -coordinate.

This is confusing, because usually means the coordinate, not an angle.
To avoid this confusion,
I often name the angle something like or instead of .
On this page, and in calculus in general, we don't use degrees as the unit of the angle.
In fact, the resulting derivative formulas don't work if your angles are represented as degrees.
Instead, we use radians, which are simply arc length on the unit circle:
an angle , as radians, means the angle that a unit circle arc of length makes.
For example, 360 degrees would be the unit circle's circumference .

Derivatives of and , with physics
Let represent time as seconds.
Let's say you are walking around in a circle of radius 1 metre so that after seconds of walking,
the distance traveled so far is metres.
In other words, your speed is 1 metre per second.

Because the arc length is the time , and because the radius of the circle is 1 (metre),
the angle on the unit circle is radians and so your location is .
The derivative of this vector is the velocity,
i.e. a vector indicating in which direction and how fast you are moving:
So to find and , for any number ,
we only need to know the and components of the velocity vector.
The velocity vector always points along the circle, because you are moving along the circle:

The length of the velocity vector is 1, because your speed is 1 (metre per second).
The length of the location vector is also 1,
because you are on the unit circle.
Looking at the picture, it is now clear that we can get the velocity vector
by rotating the location vector 90 degrees counter-clockwise.
To figure out how to rotate a vector by 90 degrees,
we first figure out how to rotate and , and then any vector
(similarly to how it's done in linear algebra).
Rotating 90 degrees counter-clockwise gives the vector .
With positive , it looks like this:

With negative , it looks like this:

Rotating 90 degrees counter-clockwise gives the vector .
With positive , the vector ends up pointing left:

With negative , the vector ends up pointing right,
but is still correct because is now positive:

Any vector rotated 90 degrees counter-clockwise is now ,
because we can rotate the and parts individually.

Let's put it all together:
We can now compare the and coordinates of the first and last vectors.
An advantage of this derivation is that it very clearly shows
why one of the derivatives has a minus in front and the other doesn't:
that comes from how 90-degree rotations work.
A disadvantage is that because it relies on physics,
it is not convincing from a purely mathematical point of view.
Derivative of
Because (TODO), we can use
the quotient rule
and plug in the derivatives of and that we just derived.
There are two ways to simplify this from here:
- Splitting the fraction into two:
- Using (a well-known trigonometry result).
Both ways to write the derivative have their own advantages and disadvantages.
You can use whichever is more convenient.
For example, is useful if you will need to compute the value of anyway,
but things like
simplify better if you write the derivative as .