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 0 answersPost by Rafael Mojica on January 28, 2015Where did you get the 6 from? at the last example in the final answer you multiplied by 6. 0 answersPost by Josh Winfield on December 17, 2012using two variables on example three is so much simpler i found. f(u)=u^2 g(x)=6sin(6x) 0 answersPost by Stephanie DahlstrÃ¶m on December 11, 2012I 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 answersPost by Gary Becraft on October 19, 2011that 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 answersPost by Atticus McCoy on October 12, 2011Awesome job I'm loving it :) 3 answersLast reply by: StefÃ¡n Berg JanssonSat Nov 26, 2011 6:31 PMPost by Abel Morales on February 19, 2011I 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.