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For more information, please see full course syllabus of College Calculus: Level I
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Lecture Comments (9)

0 answers

Post by Rafael Mojica on January 28, 2015

Where did you get the 6 from? at the last example in the final answer you multiplied by 6.

0 answers

Post by Josh Winfield on December 17, 2012

using two variables on example three is so much simpler i found. f(u)=u^2 g(x)=6sin(6x)

0 answers

Post by Stephanie Dahlström on December 11, 2012

I wonder how it works when you have a function like this: f(x) = t/(t^2+1)^0,5 I've tried the quotient rule and the chain rule but I don't get the right answer.

0 answers

Post by Gary Becraft on October 19, 2011

that was much easier to understand than the explanation given in class. I have heard others use the phrase "the stuff" and it simplifies the concept nicely.

0 answers

Post by Atticus McCoy on October 12, 2011

Awesome job I'm loving it :)

3 answers

Last reply by: Stefán Berg Jansson
Sat Nov 26, 2011 6:31 PM

Post by Abel Morales on February 19, 2011

I have a question, sin^2 (6x) wouldn't that equal cos^2 (6x), if not, why?

The Chain Rule

  • Think of the function you are trying to differentiate as a function composition.
  • You can informally think of this as involving an “outer function” and an “inner function.”
  • Some problems involve a composition of three or more functions!
  • You will be using the Chain Rule throughout the rest of calculus, so it is important to learn it really well!

The 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

The 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
  • Statement of the Chain Rule 0:09
    • Chain Rule for Three Functions
  • Lecture Example 1 1:00
  • Lecture Example 2 4:34
  • Lecture Example 3 7:23
  • Additional Example 4
  • Additional Example 5