With sine and cosine defined for all real numbers, we can now define the other 4 trig functions.
If you have trouble remembering what goes with what, there is one "co" for each reciprocal pair. Tangent and cotangent are reciprocal, sin and cosecant are reciprocal, and cosine and secant are reciprocal.
Compute their derivatives using the quotient rule. I include sine and cosine for completeness.
The arc functions are the inverse functions. They are usually indicated by a leading a. Thus atan(tan(x)) = x, and asec(sec(x)) = x, and so on.
Use the inverse function rule to find the derivatives of these arc trig functions. For instance, if y = atan(x) then x = tan(y). The derivative of x with respect to y is the derivative of tangent, or secant squared. Therefore the derivative of y with respect to x is the reciprocal, or 1 over sec2(y). But how do we get back to x? We know that x is the tangent of y, so think of y as an angle, θ if you prefer. Draw a triangle with base 1 and height x. Sure enough, x = tan(θ). The second is the hypotenuse, which equals sqrt(x2+1) (pythagorean theorem), divided by the base, which equals 1. Thus secant squared = 1+x2, and the derivative of atan(x) is 1 over 1+x2. Similar reasoning gives formulas for the other derivatives. Here they are in one table.
You've noticed that the derivatives clump together in ± pairs. Given any x, asin(x) gives one angle and acos(x) gives the other. These are complementary angles in a right triangle. Thus asin(x) + acos(x) is always 90°. Take derivatives and the right side drops to 0, hence the derivative of arcsine must be the opposite of the derivative of arccosine. Similar reasoning holds for tangent and cotanget, secant and cosecant.
Let's practice some trig differentiation. Two hallways meet at a right angle. They have widths a and b. You are carrying a long ladder on its side, and you need to turn the corner. (This is really a 2 dimensional problem; let a line segment represent the ladder.) What is the longest ladder, in terms of a and b, that will make the turn?
Pivot the ladder around the inner corner of the hallways. The shortest line corresponds to the longest ladder that will make the turn. If the horizontal hallway has width a, and the line is at angle θ to the horizontal, its length is b/c + a/s, where c and s are the cosine and sine of θ respectively. Differentiate, and set to 0, giving bs3 = ac3. The tangent of θ is the cube root of a/b. Call this tangent w. Whence w is the cube root of a/b.
Draw a right triangle whos base is 1 and altitude is w. (This gives the correct tangent.) The pythagorean theorem gives the hypotenuse, and the other trig ratios. The ladder has the following length.
b/c + a/s
(b + a/w) × sqrt(w2+1)
b⅓ × (b⅔ + a⅔) × sqrt(w2+1)
(b⅔ + a⅔) × sqrt(b⅔) × sqrt(w2+1)
(a⅔ + b⅔) × sqrt(a⅔ + b⅔)
(a⅔ + b⅔)3/2
The last formulation is symmetric in a and b, as one would expect. Set a = b and get 2×sqrt(2)×b, as one would expect. Finally, let a approach 0, and the length approaches b, as one would expect.
Let the hallways live in 3 dimensions, with a height of z. (I'm using z like the z axis, and besides, I still want to use c for cos(θ).) The ladder is still represented by a line segment.
Our line segment is turning the corner, pivoting about the inner corner of the hallway. If one end of the line is not touching the ceiling, push it up, so to speak, linearly, until it touches the ceiling. This only makes the line longer. At the same time, push the other end of the line down to the floor. Therefore we may assume, as the line turns the corner, that the leading end is always on the floor, and the trailing end is always on the ceiling.
θ is still the angle between the plane containing the line and the inner wall of the second hallway, as it was before. Turn the line, and take the minimum length as a function of θ. But computing the length of the line as a function of θ is trickier.
The projection of the line segment onto the floor has length b/c + a/s. We know this from the 2 dimensional case. The height of the ceiling is z. Apply the pythagorean theorem, and the segment has length:
sqrt((b/c + a/s)2 + z2)
We are taking the minimum of this function, and that doesn't change under square root, so consider (b/c + a/s)2 + z2. Remember that z is a constant and drops out under differentiation.
2 × (b/c + a/s) × (bs/c2 - ac/s2) = 0
One of the three factors is equal to 0. If it is the third factor, we get w = tan(θ) = cube root of a/b as before. If it is the second factor, then b/c = -a/s. The tangent is negative, and θ is out of range.
θ is as it was before. The length of the projection of the ladder onto the floor is as it was before. Same formula. That formula ended in an exponent of 3/2. It is squared by the pythagorean theorem, so we start out with:
(a⅔ + b⅔)3
Add in z2, which is a constant, then take the square root.
sqrt((a⅔ + b⅔)3 + z2)
If z = 0 we are back to our 2 dimensional formula. If z is huge then the answer approaches z. If z = a = b, square tubes meeting at a corner, the length is 3b.