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Lecture Comments (3)

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Post by Steve Denton on October 7, 2012

The above are regarding example 6.

0 answers

Post by Steve Denton on October 7, 2012

After graphing the function, it indeed slopes downward at pi.

so the slope must be negative.

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Post by Steve Denton on October 7, 2012

I got negative 1 / pi squared on my calculator.

Quotient Rule

  • Organization is vital
  • Write more, more room to think, make less mistakes
  • Careful with distribution of “-” sign when evaluating

Quotient Rule

Given f(x) = 3x and g(x) = 4x, find [d/dx] [f(x)/g(x)]
  • [d/dx] [f(x)/g(x)] = [(g(x)f′(x) − f(x)g′(x))/(g(x)2)]
  • = [(4x [d/dx] 3x − 3x [d/dx] 4x)/((4x)2)]
  • = [(12x − 12x)/(16x2)]
0
Find f′(x) if f(x) = [3x/(x2 + 1)]
  • f′(x) = [3x/(x2 + 1)]
  • = [((x2 + 1) [d/dx] 3x − 3x [d/dx] (x2 + 1))/((x2 + 1)2)]
  • = [((x2 + 1)3 − 3x(2x))/((x2 + 1)2)]
  • = [(3x2 −6x2 + 3)/((x2 + 1)2)]
[(−3 (x2 − 1))/((x2 + 1)2)]
Find the derivative of tan(x) using the quotient rule
  • [d/dx] tan(x) = [d/dx] [sin(x)/cos(x)]
  • = [(cos(x) [d/dx] sin(x) − sin(x) [d/dx] cos(x))/(cos2(x))]
  • = [(cos(x)cos(x) − sin(x) (−sin(x)))/(cos2(x))]
  • = [(cos2(x) + sin2(x))/(cos2(x))]
  • = [1/(cos2(x))]
sec2(x)
Find the derivative of cot(x) using the quotient rule
  • [d/dx] cot(x) = [d/dx] [cos(x)/sin(x)]
  • = [(sin(x) [d/dx] cos(x) − cos(x) [d/dx] sin(x))/(sin2(x))]
  • = [(sin(x) (−sin(x)) − cos(x) cos(x))/(sin2(x))]
  • = −[(sin2(x) + cos2(x))/(sin2(x))]
  • = −[1/(sin2(x))]
−csc2(x)
Find the derivative of sec(x) using the quotient rule
  • [d/dx] sec(x) = [d/dx] [1/cos(x)]
  • = [(cos(x) [d/dx] 1 − 1 [d/dx] cos(x))/(cos2(x))]
  • = [(0 − (−sin(x)))/(cos2(x))]
  • = [sin(x)/(cos2(x))]
  • = sec(x) [sin(x)/cos(x)]
sec(x) tan(x)
Find f′(x) if f(x) = [(x3)/((5x + 4)tan(x))]
  • f′(x) = [((5x + 4) tan(x) [d/dx] x3 − x3 [d/dx] ((5x + 4) tan(x)t))/(t((5x + 4)tan(x))2)]
  • = [((5x + 4) tan(x) (3x2) − x3 [d/dx] ((5x + 4) tan(x)))/((5x + 4)2 tan2(x))]
  • = [((5x + 4) tan(x) (3x2) − x3 ( (5x + 4) [d/dx] tan(x) + tan(x) [d/dx] (5x + 4) ))/((5x + 4)2 tan2(x))]
  • = [((5x + 4) tan(x) (3x2) − x3 ( (5x + 4) sec2(x) + tan(x) 5 ))/((5x + 4)2 tan2(x))]
[((5x + 4) tan(x) (3x2) − x3 ( (5x + 4) sec2(x) + 5tan(x)))/((5x + 4)2 tan2(x))]
Find f′(x) if f(x) = [(√x + 3)/(x4 −16)]
  • f′(x) = [(√x + 3)/(x4 − 16)]
  • = [((x4 − 16) [d/dx] (√x + 3) − (√x + 3) [d/dx] (x4 − 16))/((x4 − 16)2)]
[((x4 − 16) [1/2] x−[1/2] − (√x + 3) (4x3))/((x4 − 16)2)]
Find f′(x) if f(x) = [(.2x6)/(.1x + cos(x))]
  • f′(x) = [((.1x + cos(x)) [d/dx] .2x6 − (.2x6) [d/dx] (.1x + cos(x)))/((.1x + cos(x))2)]
[((.1x + cos(x)) 1.2x5 − .2x6 (.1 − sin(x)))/((.1x + cos(x))2)]
Find f′(t) if f(t) = [(t2)/sin(t)]
  • The letter or symbol used for the variable is not important. What's important is consistency.
  • f′(t) = [(sin(t) [d/dt] t2 − t2 [d/dt] sin(t))/(sin2(t))]
  • = [(sin(t) (2t) − t2 cos(t))/(sin2(t))]
[(t(2sin(t) − t cos(t)))/(sin2(t))]
Find f′(1) if f(z) = [(z − 3)/(2 − z)]
  • f′(t) = [((2 − z) [d/dz] (z − 3) − (z − 3) [d/dz] (2 − z))/((2 − z)2)]
  • = [((2 − z) 1 − (z − 3) (−1))/((2 − z)2)]
  • = [(2 − z + z − 3)/((2 − z)2)] =
−[1/((2 − z)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

Quotient 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
  • Quotient Rule 0:07
    • Definition
    • Example 1
  • Example 2: With No X In Numerator 2:49
  • Example 3 4:30
  • Example 4: With Decimals 6:46
  • Example 5 8:53
  • Example 6: With Trig Functions 12:55