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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Related Books

### Higher Derivatives

- This topic simply involves repeatedly differentiating!
- Learn the two different types of notation for higher derivatives – both are used a lot!
- Some problems will ask you to find a formula for the
*n*th derivative – in this case, differentiate enough times to see the pattern and then build a formula describing the pattern. - Some problems will ask you to find, for example, the 100
^{th}derivative – in this case, again differentiate enough times to see the pattern and then figure out the result!

### Higher Derivatives

Find the second derivative of f(t) = 3t

^{2}+ 11t + 17- f′(t) = [d/dt] (3t
^{2}+ 11t + 17) - = 6t + 11
- f"(t) = [d/dt] (6t + 11)

6

Find the third derivative of f(t) = 3t

^{2}+ 11t + 17- f"′(t) = [d/dt] f"(t)
- = [d/dt] (6)

0

Find the second derivative of f(x) = tan(3x)

- f′(x) = [d/dx] tan(3x)
- = sec
^{2}(3x) [d/dx] (3x) - = 3 sec
^{2}(3x) - f"(x) = [d/dx] 3 sec
^{2}(3x) - u = sec(3x), u′ = sec(3x)tan(3x) [d/dx] 3x = 3 sec(3x)tan(3x)
- f"(x) = 3 [d/dx] u
^{2} - = 6u u′
- = 6 sec(3x) (3 sec(3x)tan(3x))

18 sec

^{2}(3x) tan(3x)Find the second derivative of f(x) = x − [(x

^{3})/6] + [(x^{5})/120]- f′(x) = [d/dx] (x − [(x
^{3})/6] + [(x^{5})/120]) - = 1 − [(3x
^{2})/6] + [(5x^{4})/120] - = 1 − [(x
^{2})/2] + [(x^{4})/24] - f"(x) = [d/dx] (1 − [(x
^{2})/2] + [(x^{4})/24]) - = 0 − [2x/2] + [(4x
^{3})/24]

−x + [(x

f(x) here might look familiar. It is a partial series representation of sin(x)

^{3})/6]f(x) here might look familiar. It is a partial series representation of sin(x)

Find the second derivative of f(t) = t

^{2}sin(5t)- f′(x) = [d/dt] (t
^{2}sin(5t)) - = t
^{2}[d/dt] sin(5t) + sin(5t) [d/dx] t^{2} - = t
^{2}(5cos(5t)) + sin(5t) (2t) - = 5 t
^{2}cos(5t) + 2t sin(5t) - f"(x) = [d/dt] (5 t
^{2}cos(5t) + 2t sin(5t)) - = 5t
^{2}[d/dt] cos(5t) + 5cos(5t) [d/dt] t^{2}+ 2t [d/dt] sin(5t) + 2 sin(5t) [d/dt] t - = −25t
^{2}sin(5t) + 10t cos(5t) + 10t cos(5t) + 2sin(5t)

−25t

^{2}sin(5t) + 20t cos(5t) + 2sin(5t)Find the second derivative of x(t) = 5 cos(2t − [(π)/4])

- x′(t) = [d/dt] (5 cos(2t − [(π)/4]) )
- = −5 sin(2t − [(π)/4]) [d/dt] (2t − [(π)/4])
- = −10 sin(2t − [(π)/4])
- x"(t) = [d/dt] (−10 sin(2t − [(π)/4]))
- = −10 cos(2t − [(π)/4]) [d/dt] (2t − [(π)/4]) =
- x(t) in this problem represents an example of simple harmonic motion, which can be used to describe the motion of things like springs

−20 cos(2t − [(π)/4])

Find the second derivative of f(x) = [1/(x

^{2})]- f(x) = [1/(x
^{2})] - = x
^{−2} - f′(x) = [d/dx] x
^{−2} - = −2x
^{−3} - f"(x) = [d/dx] (−2x
^{−3}) - = −2 [d/dx] x
^{−3} - = −2(−3x
^{−4}) - = 6x
^{−4}=

[6/(x

^{4})]Find the second derivative of f(x) = √{x

^{3}+ 5x}- f′(x) = [d/dx] √{x
^{3}+ 5x} - u = x
^{3}+ 5x, u′ = 3x^{2}+ 5, u" = 6x - f′(x) = [d/dx] u
^{[1/2]} - = [1/2] u
^{−[1/2]}u′ - f"(x) = [d/dx] [1/2] u
^{−[1/2]}u′ - = [1/2] ( u
^{−[1/2]}[d/dx] u′+ u′[d/dx] (u^{−[1/2]})) - = [1/2] ( u
^{−[1/2]}u" + u′(−[1/2] u^{−[3/2]}u′)) - = [1/2] ( u
^{−[1/2]}u" − [1/2] (u′)^{2}u^{−[3/2]})

[1/2] ( (x

^{3}+ 5x)^{−[1/2]}(6x) − [1/2] (3x^{2}+ 5)^{2}(x^{3}+ 5x)^{−[3/2]})Find the 2nd, 4th, 6th, and 8th derivatives of y = sin(x)

- Here's another way to write higher order derivatives
- [d/dx] [d/dx]z = [(d
^{2})/(dx^{2})] z (2nd derivative of z with respect to x) - [(d
^{3})/(dx^{3})] (f(x) equivalent to f"′(x)) - [(d
^{9}y)/(dx^{9})] (9th derivative of y with respect to x)

y = sin(x)

dy = cos(x)

[d/dy]2 = −sin(x)

[d/dy]3 = −cos(x)

[d/dy]4 = sin(x)

[d/dy]5 = cos(x)

[d/dy]6 = −sin(x)

[d/dy]7 = −cos(x)

[d/dy]8 = sin(x)

dy = cos(x)

[d/dy]2 = −sin(x)

[d/dy]3 = −cos(x)

[d/dy]4 = sin(x)

[d/dy]5 = cos(x)

[d/dy]6 = −sin(x)

[d/dy]7 = −cos(x)

[d/dy]8 = sin(x)

Find the 2nd, 4th, 6th, and 8th derivatives of y = cos(2x)

y = cos(2x)

dy = −2sin(2x)

[d/dy]2 = −4cos(2x)

[d/dy]3 = 8sin(2x)

[d/dy]4 = 16cos(2x) = 2

[d/dy]6 = −(2

[d/dy]8 = (2

dy = −2sin(2x)

[d/dy]2 = −4cos(2x)

[d/dy]3 = 8sin(2x)

[d/dy]4 = 16cos(2x) = 2

^{4}cos(2x)[d/dy]6 = −(2

^{6})cos(2x)[d/dy]8 = (2

^{8})cos(2x)*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Higher Derivatives

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Notation 0:08
- First Type
- Second Type
- Lecture Example 1 1:41
- Lecture Example 2 3:15
- Lecture Example 3 4:57
- Additional Example 4
- Additional Example 5

0 answers

Post by Amina Kalla on February 16, 2013

how do we find the nth derivatives of Sin^2(x)