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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.
































0 answers
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)?
0 answers
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
0 answers
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?
0 answers
Post by Nicolette Reilly on November 17, 2011
how do you find the derivative of f(x)=cos (pi/x) ?
0 answers
Post by Stefán 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
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.