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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:

Differentiation Interactive Applet - trigonometric functions.

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).

Answer

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`.

Answer

` 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.

Answer

`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`

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Continues below

Exercises

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

Answer

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)`

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

Answer

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`

Easy to understand math videos:
MathTutorDVD.com

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

Answer

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)`

Adding

`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`

Adding

`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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