2.3, Derivatives of Trig Functions

We take an extra class day after demonstrating the product and quotient rules for students to practice those new rules and to demonstrate how the quotient rule leads to the derivatives of trig functions.

Once we understand the quotient rule, it is easy to take the derivatives of

1. tan x = (sin x)/(cos x)
2. sec x = 1/(cos x)
3. cot x = (cos x)/(sin x)
4. csc x = 1/(sin x)

In each case we can use the quotient rule and possibly the Pythagorean identity

cos^2 x + sin^2 x = 1.

For example, in taking the derivative of tan x = (sin x)/(cos x), by the quotient rule, we get the numerator

(cos x)(cos x)-(sin x)(-sin x) = (cos x)^2 + (sin x)^2 = 1

while the denominator is (cos x)^2.

So our answer is 1/(cos x)^2 which is the same as (sec x)^2.

It is customary in the notation for trig functions to put the exponent over the trig function, writing sec^2 x for (sec x)^2 and so the derivative of tan x is sec^2 x.

Using the quotient rule, the derivative of sec x = 1/(cos x) will be  sin x/(cos x)^2 (since the derivative of 1 is 0) and it is customary to rewrite
sin x/(cos x)^2 = (sin x/cos x)(1/cos x) = tan x sec x.

Similar steps give formulas for the derivatives of the cotangent and cosecant functions.

Here is a summary.

One might notice that the trig functions which have minus signs in the answer are all "co"-functions. There is a good reason for this!  We will look more at this when we explore the Chain Rule.