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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of Calculus AB
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Lecture Comments (6)

0 answers

Post by Quinn Hollister on May 23, 2015

I believe 2nd derivative is acceleration, and first derivative is the velocity function

0 answers

Post by Nolan Zhang on April 10, 2015

example 3 was quite long, you should probably use another one

0 answers

Post by Quazi Niloy on October 13, 2014

On example 3, wouldn't it have been simple to take sin(t)/t^3, and then just multiply it where you have sin(t)*(t^-3) and then apply either the product or the chain rule?

2 answers

Last reply by: John Zhu
Tue Aug 13, 2013 8:36 PM

Post by Vanessa Munoz on June 12, 2013

I don't understand in 10:17, how you can replace with zero if it makes de denominator zero..wouldn't that be undefined?

Higher Order Derivatives

  • Different order derivatives have different practical meanings
  • To avoid confusion: treat each level of derivative as brand new derivative
  • Higher order derivatives are easier to solve because of eliminated terms

Higher Order Derivatives

Find the second derivative of f(t) = 3t2 + 11t + 17
  • f′(t) = [d/dt] (3t2 + 11t + 17)
  • = 6t + 11
  • f"(t) = [d/dt] (6t + 11)
6
Find the third derivative of f(t) = 3t2 + 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)
  • = sec2(3x) [d/dx] (3x)
  • = 3 sec2(3x)
  • f"(x) = [d/dx] 3 sec2(3x)
  • u = sec(3x), u′ = sec(3x)tan(3x) [d/dx] 3x = 3 sec(3x)tan(3x)
  • f"(x) = 3 [d/dx] u2
  • = 6u u′
  • = 6 sec(3x) (3 sec(3x)tan(3x))
18 sec2(3x) tan(3x)
Find the second derivative of f(x) = x − [(x3)/6] + [(x5)/120]
  • f′(x) = [d/dx] (x − [(x3)/6] + [(x5)/120])
  • = 1 − [(3x2)/6] + [(5x4)/120]
  • = 1 − [(x2)/2] + [(x4)/24]
  • f"(x) = [d/dx] (1 − [(x2)/2] + [(x4)/24])
  • = 0 − [2x/2] + [(4x3)/24]
−x + [(x3)/6]
f(x) here might look familiar. It is a partial series representation of sin(x)
Find the second derivative of f(t) = t2 sin(5t)
  • f′(x) = [d/dt] (t2 sin(5t))
  • = t2 [d/dt] sin(5t) + sin(5t) [d/dx] t2
  • = t2 (5cos(5t)) + sin(5t) (2t)
  • = 5 t2 cos(5t) + 2t sin(5t)
  • f"(x) = [d/dt] (5 t2 cos(5t) + 2t sin(5t))
  • = 5t2 [d/dt] cos(5t) + 5cos(5t) [d/dt] t2 + 2t [d/dt] sin(5t) + 2 sin(5t) [d/dt] t
  • = −25t2sin(5t) + 10t cos(5t) + 10t cos(5t) + 2sin(5t)
−25t2sin(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/(x2)]
  • f(x) = [1/(x2)]
  • = 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/(x4)]
Find the second derivative of f(x) = √{x3 + 5x}
  • f′(x) = [d/dx] √{x3 + 5x}
  • u = x3 + 5x, u′ = 3x2 + 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] ( (x3 + 5x)−[1/2] (6x) − [1/2] (3x2 + 5)2 (x3 + 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 = [(d2)/(dx2)] z (2nd derivative of z with respect to x)
  • [(d3)/(dx3)] (f(x) equivalent to f"′(x))
  • [(d9 y)/(dx9)] (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)
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) = 24 cos(2x)
[d/dy]6 = −(26)cos(2x)
[d/dy]8 = (28)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 Order 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
  • Types of Derivatives: Higher Order Derivatives 0:07
    • 1st Derivative / F Prime
    • 2nd Derivative
    • 3rd Derivative
    • Example 1
  • Example 2: Find 3rd Derivative 3:13
  • Example 3: Acceleration 4:25
  • Example 4 10:20
  • Example 5: 2nd Derivative 12:11