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`
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)`
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`
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)`