# 2. Derivatives of Csc, Sec and Cot Functions

by M. Bourne

By using the quotient rule and trigonometric identities, we can obtain the following derivatives:

(d(csc x))/(dx)=-csc x cot x

(d(sec x))/(dx)=sec x tan x

(d(cot x))/(dx)=-csc^2 x

In words, we would say:

The derivative of csc x is -csc x cot x,
The derivative of sec x is sec x tan x and
The derivative of cot x is -csc^2 x.

Explore animations of these functions with their derivatives here:

If u = f(x) is a function of x, then by using the chain rule, we have:

(d(csc u))/(dx)=-csc u\ cot u(du)/(dx)

(d(sec u))/(dx)=sec u\ tan u(du)/(dx)

(d(cot u))/(dx)=-csc^2u(du)/(dx)

### Example 1

Find the derivative of s = sec(3t + 2).

### Example 2

Find the derivative of x = θ^3 csc 2θ.

### Example 3

Find the derivative of y = sec43x.

Continues below

## Exercises

1. Find the derivative of y = csc2(2x2).

2. Find the derivative of y = sec2 2x.

3. Find the derivative of 3 cot(x + y) = cos y2.

top

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

### Calculus Lessons on DVD

Easy to understand calculus lessons on DVD. See samples before you commit.