For more information, please see full course syllabus of College Calculus: Level I

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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

### The Quotient Rule

- Memorize the Quotient Rule – you will be using it a lot!
- A constant factor can be factored out in front of a derivative.
- A function of the form can be re-written as if you prefer the Chain Rule to the Quotient Rule.
- Avoid common errors by remembering to use the Quotient Rule whenever you are differentiating a quotient!

### The 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)/(16x
^{2})]

0

Find f′(x) if f(x) = [3x/(x

^{2}+ 1)]- f′(x) = [3x/(x
^{2}+ 1)] - = [((x
^{2}+ 1) [d/dx] 3x − 3x [d/dx] (x^{2}+ 1))/((x^{2}+ 1)^{2})] - = [((x
^{2}+ 1)3 − 3x(2x))/((x^{2}+ 1)^{2})] - = [(3x
^{2}−6x^{2}+ 3)/((x^{2}+ 1)^{2})]

[(−3 (x

^{2}− 1))/((x^{2}+ 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))/(cos
^{2}(x))] - = [(cos(x)cos(x) − sin(x) (−sin(x)))/(cos
^{2}(x))] - = [(cos
^{2}(x) + sin^{2}(x))/(cos^{2}(x))] - = [1/(cos
^{2}(x))]

sec

^{2}(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))/(sin
^{2}(x))] - = [(sin(x) (−sin(x)) − cos(x) cos(x))/(sin
^{2}(x))] - = −[(sin
^{2}(x) + cos^{2}(x))/(sin^{2}(x))] - = −[1/(sin
^{2}(x))]

−csc

^{2}(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))/(cos
^{2}(x))] - = [(0 − (−sin(x)))/(cos
^{2}(x))] - = [sin(x)/(cos
^{2}(x))] - = sec(x) [sin(x)/cos(x)]

sec(x) tan(x)

Find f′(x) if f(x) = [(x

^{3})/((5x + 4)tan(x))]- f′(x) = [((5x + 4) tan(x) [d/dx] x
^{3}− x^{3}[d/dx] ((5x + 4) tan(x)t))/(t((5x + 4)tan(x))^{2})] - = [((5x + 4) tan(x) (3x
^{2}) − x^{3}[d/dx] ((5x + 4) tan(x)))/((5x + 4)^{2}tan^{2}(x))] - = [((5x + 4) tan(x) (3x
^{2}) − x^{3}( (5x + 4) [d/dx] tan(x) + tan(x) [d/dx] (5x + 4) ))/((5x + 4)^{2}tan^{2}(x))] - = [((5x + 4) tan(x) (3x
^{2}) − x^{3}( (5x + 4) sec^{2}(x) + tan(x) 5 ))/((5x + 4)^{2}tan^{2}(x))]

[((5x + 4) tan(x) (3x

^{2}) − x^{3}( (5x + 4) sec^{2}(x) + 5tan(x)))/((5x + 4)^{2}tan^{2}(x))]Find f′(x) if f(x) = [(√x + 3)/(x

^{4}−16)]- f′(x) = [(√x + 3)/(x
^{4}− 16)] - = [((x
^{4}− 16) [d/dx] (√x + 3) − (√x + 3) [d/dx] (x^{4}− 16))/((x^{4}− 16)^{2})]

[((x

^{4}− 16) [1/2] x^{−[1/2]}− (√x + 3) (4x^{3}))/((x^{4}− 16)^{2})]Find f′(x) if f(x) = [(.2x

^{6})/(.1x + cos(x))]- f′(x) = [((.1x + cos(x)) [d/dx] .2x
^{6}− (.2x^{6}) [d/dx] (.1x + cos(x)))/((.1x + cos(x))^{2})]

[((.1x + cos(x)) 1.2x

^{5}− .2x^{6}(.1 − sin(x)))/((.1x + cos(x))^{2})]Find f′(t) if f(t) = [(t

^{2})/sin(t)]- The letter or symbol used for the variable is not important. What's important is consistency.
- f′(t) = [(sin(t) [d/dt] t
^{2}− t^{2}[d/dt] sin(t))/(sin^{2}(t))] - = [(sin(t) (2t) − t
^{2}cos(t))/(sin^{2}(t))]

[(t(2sin(t) − t cos(t)))/(sin

^{2}(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

### The 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
- Statement of the Quotient Rule 0:07
- Carrying out the Differentiation
- Quotient Rule in Words
- Lecture Example 1 1:19
- Lecture Example 2 4:23
- Lecture Example 3 8:00
- Additional Example 4
- Additional Example 5

0 answers

Post by Mohamed Al Mohannadi on September 11, 2016

We can further simply "Lecture Example 1."

We can write -16 / (6x-5)^2 as - [ 4 / 6x - 5 ] ^ 2

Am I right?

0 answers

Post by Ashley Simon on May 5, 2015

at 3:38, why is 6 negative??

0 answers

Post by Chateau Siqueira on September 3, 2013

Where Can I find a Lecture about the Difference quotient? Thanks!

1 answer

Last reply by: Martina Alvarez

Tue Dec 6, 2011 1:20 PM

Post by naman ahmad on October 24, 2011

on the last additional problem, last step we had:(e^x+7)-(Xe^x)/(e^x+7)^2. we could have cancelled out the e^x+7 and remained with -Xe^x/e^x+7.

1 answer

Last reply by: StefÃ¡n Berg Jansson

Fri Nov 25, 2011 7:10 PM

Post by Xiaosong Gao on January 17, 2011

example 3 can be further simplified to: 3^(1/2)x/1+x^3

1 answer

Last reply by: Larry Davis

Wed Sep 15, 2010 10:25 AM

Post by Larry Davis on September 15, 2010

on ex 2 you used the wrongterms in the numerator. it should have been ex d/dx[X8]-

X8 d/dx[ex]/[x8]2 gives you =ex(8-x)/X9