# 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).

Put u = 3t + 2. Then:

s=sec u and

(du)/(dt) = 3

So using Chain Rule, we have:

(ds)/(dt) =(ds)/(du) (du)/(dt)

=sec(u) tan (u)(3)

=3 sec(3t+2) tan (3t+2)

### Example 2

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

 x=theta^3csc 2 theta

If we let u=theta^3 and v=csc 2 theta, then

(dx)/(d theta) =u(dv)/(d theta)+v (du)/(d theta)

=theta^3(-csc 2 theta cot 2 theta)(2)+ csc 2 theta(3 theta^2)

=theta^2(csc 2 theta)(-2 theta cot 2 theta+3)

### Example 3

Find the derivative of y = sec43x.

y=sec^4 3x

Let y=u^4, where u=sec 3x.

Then

(dy)/(dx)=(dy)/(du)(du)/(dx)

=4u^3(sec 3x tan 3x)(3)

=4(sec^3 3x)(sec 3x tan 3x)(3)

=12 sec^4 3x tan 3x

Continues below

## Exercises

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

This is an example of a function of a function of a function, and we need to apply chain rule 3 times.

Let u = 2x2 and v = csc u.

So y = v2

 (dy)/(dx)=(dy)/(dv)(dv)/(du)(du)/(dx)

=[2v][-csc u cot u](4x)

=[2 csc(2x^2)] xx[(-csc 2x^2)(cot 2x^2)](4x)

=-8x(csc^2 2x^2)(cot 2x^2)

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

This is also an example of a function of a function of a function, and we need to apply chain rule 3 times.

This can be written y = sec^2u where u = 2x.

If we let v = sec u then y=v^2.

So we have:

y=v^2=sec^2 u

Then

(dy)/(dx)=(dy)/(dv)(dv)/(du)(du)/(dx)

=(2v)(sec u tan u)(2)

=(4v)(sec u tan u)

=(4 sec u)(sec u tan u)

=4 sec^2 u tan u

=4 sec^2 2x tan 2x

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

This is an implicit function.

3 cot(x + y) = cos y2

For the left hand side, we put u = x + y.

Differentiating 3 cot u gives us:

3(-csc^2 u)((du)/(dx))

Substituting for u and performing the (du)/(dx) part gives us:

-3 csc^2(x+y)(1+(dy)/(dx))

On the right hand side, we let u = y2. Differentiating cos u gives us:

(-sin u)((du)/(dx))

Substituting for u and performing the (du)/(dx) part gives us:

(-sin y^2)(2y(dy)/(dx))

We put both sides together:

-3 csc^2(x+y)(1+(dy)/(dx)) =(-sin y^2)(2y(dy)/(dx))

Expanding gives:

-3 csc^2(x+y) -3 csc^2(x+y)(dy)/(dx) =-2y sin y^2(dy)/(dx)

2y sin y^2(dy)/(dx)

to both sides:

-3 csc^2(x+y) -3 csc^2(x+y)(dy)/(dx) +2y sin y^2(dy)/(dx) =0

3 csc^2(x+y)

to both sides:

-3 csc^2(x+y)(dy)/(dx) +2y sin y^2(dy)/(dx) =3 csc^2(x+y)

Factoring out the dy/dx term:

[2y sin y^2-3 csc^2(x+y)](dy)/(dx) =3 csc^2(x+y)

This gives us:

(dy)/(dx) =(3 csc^2(x+y))/(2y sin y^2 -\ 3 csc^2(x+y)

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