By making use of the quotient rule and also trigonometric identities, we can achieve the adhering to 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 certainly say:
The derivative of `csc x` is `-csc x cot x`, The derivative of `sec x` is `sec x tan x` and also The derivative the `cot x` is `-csc^2 x`.
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Differentiation interaction Applet - trigonometric functions.
If u = f(x) is a role of x, then by utilizing the chain rule, us 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, us have:
`(ds)/(dt) =(ds)/(du) (du)/(dt)`
`=sec(u) tan (u)(3)`
`=3 sec(3t+2) tan (3t+2)`
Example 2
Find the derivative that `x = θ^3 csc 2θ`.
Answer
` x=theta^3csc 2 theta `
If us let `u=theta^3` and also `v=csc 2 theta`, climate
`(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`, wherein `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. Discover the derivative the y = csc2(2x2).
Answer
This is an example of a function of a role of a function, and also we require to apply chain dominance 3 times.
Let u = 2x2 and also 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 additionally an example of a duty of a role of a function, and also we require to use chain dominion 3 times.
This have the right to be composed `y = sec^2u` wherein `u = 2x`.
If us let `v = sec u` climate `y=v^2`.
So us 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. Discover the derivative of 3 cot(x + y) = cos y2.
Answer
This is an latent function.
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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 because that `u` and performing the `(du)/(dx)` part gives us:
`-3 csc^2(x+y)(1+(dy)/(dx))`
On the ideal hand side, us let u = y2. Separating `cos u` gives us:
`(-sin u)((du)/(dx))`
Substituting because that `u` and performing the `(du)/(dx)` part offers us:
`(-sin y^2)(2y(dy)/(dx))`
We placed both political parties 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 provides us:
`(dy)/(dx)` `=(3 csc^2(x+y))/(2y sin y^2` `- 3 csc^2(x+y)`
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