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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (12)

0 answers

Post by Gautham Padmakumar on November 1, 2015

You actually forgot to do two parts of example 2. In the quotient rule you were supposed to multiply by x and also forgot to differentiate the inner function (2+x^2)

0 answers

Post by Nolan Zhang on April 10, 2015

that explanation of Chain rule really helped me a lot!

0 answers

Post by John Michael Musaazi on September 8, 2014

in the first example were do you get the 2/3

1 answer

Last reply by: Andrew Mu
Tue Jul 9, 2013 10:36 AM

Post by Steve Denton on October 11, 2012

At 16:40, is the derivative of cos^2(2x)= -2sin(2x) or -2sin(x)?

1 answer

Last reply by: John Zhu
Mon Aug 12, 2013 9:07 PM

Post by Steve Denton on October 8, 2012

You're going to show us the easier way, I hope. (ex,2-4)

4 answers

Last reply by: Gautham Padmakumar
Sun Nov 1, 2015 12:22 PM

Post by liqun zhu on June 2, 2012

At 8:48, don't you have to multiply d(u)/d(x) to b'? I think you forgot the g'(x) part of the function F'(x) = f(g(x)) g'(x)

Chain Rule

  • Or,
  • Don’t forget to multiply after evaluating (common hasty error)

Chain Rule

Find f′(x) if f(x) = (3x5 + 7x + 11)13
  • u(x) = 3x5 + 7x + 11
  • [du/dx] = u′(x) = 15x4 + 7
  • f(x) = u13
  • f′(x) = [d/dx] u13
  • f′(x) = 13 u12 [du/dx]
13(3x5 + 7x + 11)12 (15x4 + 7)
Find f′(x) if f(x) = √{x3 + 5x}
  • u = x3 + 5x
  • [du/dx] = 3x2 + 5
  • f′(x) = [d/dx] √u
  • f′(x) = [d/dx] u[1/2]
  • = [1/2] (u−[1/2]) [du/dx]
  • = [du/(2√u)] =
[(3x2 + 5)/(2√{x3 + 5x})]
Find f′(x) if f(x) = sin2(x)
  • u = sin(x)
  • u′ = cos(x)
  • f′(x) = [d/dx] u2
  • = 2 u u′
  • = 2 sin(x) cos(x) =
sin(2x)
Find f′(x) if f(x) = (sec(5x))9
  • u = sec(5x), v = 5x, v′ = 5
  • u = sec(v)
  • u′ = sec(v)tan(v) v′
  • u′ = 5sec(5x)tan(5x)
  • f′(x) = [d/dx] u9
  • = 9 u8 u′
  • = 9 (sec(5x))8 (5 sec(5x) tan(5x)) =
45 tan(5x) (sec(5x))9
Find f′(x) if f(x) = cot(5x2 + 3x)
  • f′(x) = −csc2(5x2 + 3x) [d/dx] (5x2 + 3x)
  • = −csc2(5x2 + 3x) (10x + 3) =
−(10x + 3)csc2(5x2 + 3x)
Find f′(x) if f(x) = [(x2)/tan(3x)]
  • g = x2, g′ = 2x
  • h = tan(3x), h′ = sec2(3x) [d/dx] 3x = 3 sec2(3x)
  • f′(x) = [d/dx] [g/h]
  • = [(h g′− g h′)/(h2)]
  • = [(tan(3x) (2x) − x2 (3 sec2(3x)))/(tan2(3x))] =
[(2x tan(3x) − 3x2 sec2(3x))/(tan2(3x))]
Find f′(x) if f(x) = (x2 + 8x + 16)(x + 4)3
  • f(x) = (x2 + 8x + 16) (x + 4)3
  • = (x + 4)2 (x + 4)3
  • = (x + 4)5
  • f′(x) = 5 (x + 4)4 [d/dx] (x + 4)
5 (x + 4)4
Find f′(x) if f(x) = (7x9 + √{3x})(x + 1)4
  • u = 7x9 + √{3x}, u′ = 63x8 + [(√3)/(2√x)]
  • v = (x + 1)4, v′ = 4 (x + 1)3 [d/dx] (x + 1) = 4 (x + 1)3
  • f′(x) = [d/dx] u v
  • = u v′+ v u′=
(7x9 + √{3x})(4 (x + 1)3) + (x + 1)4 (63x8 + [(√3)/(2√x)])
Find f′(x) if f(x) = sin(sin(x))
  • u = sin(x), u′ = cos(x)
  • f′(x) = [d/dx] sin(u)
  • = cos(u) u′=
cos(sin(x)) cos(x)
Find f′(x) if f(x) = √{csc(11x)}
  • u = csc(11x), v = 11x, v′ = 11
  • u = csc(v), u′ = −cot(v)csc(v) v′
  • u′ = −11 cot(11x)csc(11x)
  • f′(x) = [d/dx] u[1/2]
  • = [1/(2√u)] u′
  • = [(−11 cot(11x)csc(11x))/(2√{csc(11x)})] =
[(−11cot(11x)√{csc(11x)})/2]

*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

Chain Rule

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
  • Chain Rule 0:07
    • Definition
    • Example 1: Applying the Chain Rule
  • Example 2 4:25
  • Example 3 6:02
  • Example 4 9:25
  • Example 5 12:47
  • Example 6 15:27