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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

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### Trigonometric Derivatives

- You should memorize these formulas – they are important!
- The derivation of the results for and is famous and involves a couple famous limits.
- One type of problem here simply incorporates trigonometric functions into differentiation problems involving, for example, the Chain Rule.
- You can use the derivative formulas for and to derive the derivative formulas for the other trigonometric functions.

### Trigonometric Derivatives

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
- Six Basic Trigonometric Functions 0:11
- Patterns
- Lecture Example 1 1:18
- Lecture Example 2 7:38
- Lecture Example 3 12:15
- Lecture Example 4 14:25
- Additional Example 5
- Additional Example 6

0 answers

Post by abbas esmailzadeh on December 5, 2011

trigonometric derivatives ex,4

in step 3

x[-x csc(x)cot(x)-2csc(x)/x^4

what is the opening bracketts after the first x

and what is the( - )sign is after the first opening bracketts.

0 answers

Post by StefÃ¡n Berg Jansson on November 25, 2011

I try to help out, no promises tho... It will be good practice for me.

@Ann Cea

When she factored out x. If we think about it seperatly and factor only csc(x)*2x, we get the following.

csc(x)*2x = csc(x)*2*x = x(csc(x)*2) =

x(2csc(x))

@Ron Weldy on February 25 at 09:21:47 AM

If it is only plugging in x then,

y=1+2sinx;x=pie/6

y=1+2sin(pi/6)

y=1+2*0,5(sin(pi/6) = 0,5)

y=1+1 = 2

If you were suppose to take the derivative first and then plug in x:

y'=2cos(x) (1' = 0 and 2sin' = 2cos)

Plug in pi/6 where x is

2cos(pi/6)= 2*(sqrt(3)/2) = sqrt(3)

@Nicolette Reilly

f(x)=cos(pi/x)

I will use the cain and the quotient rule.

First I will use the cain rule and get:

cos'(pi/x)*(pi/x)' = -sin(pi/x)*(pi/x)'

Then I will use the quotient rule on (pi/x)'

and then multiply it with -sin(pi/x)

-sin(pi/x)*(0*x-pi*1)/x^2=

(-sin(pi/x)*-pi)/x^2 =

sin(pi/x)pi/x^2

Hope I did not make any mistakes.. I often do

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Post by Nicolette Reilly on November 17, 2011

how do you find the derivative of f(x)=cos (pi/x) ?

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Post by Ron Weldy on March 2, 2011

Where do we go to ask questions? They said we could ask questions and get help is that true?

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Post by Ron Weldy on February 25, 2011

I have a problem that looks like

y=1+2sinx;x=pie/6

I'm not sure what to do

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Post by Ann Cea on December 27, 2010

On example IV, where did the x go on the fourth step of the equation where it went from csc(x)2x to 2csc(x)?