I am having a real problem with differentiating trig functions. I have searched for a reasonable answer and keep coming up with more of the same. The trig class I took as a prerequisite did not stress the parentheses involved with trig. I think the problem is that I don't understand how the parentheses make a difference. I am at a point where I can't move on without this part of the puzzle. I have to differentiate with the chain rule for trig functions and am lost!!!!
Perhaps you could post a problem.
You need to just remember the derivatives of the trig functions
y=sin(x) y'(x)=cos(x) no chain rule needed
y=sin(2x) y'(x)=2cos(2x) chain rule needed because you are dealing with a composition of functions inner function is (2x) and outer function is sin(x)
The problem is that my proffessor requires that you solve the problem as directed so when it says use the chain rule you must. I have a hard time seeing when there is a "nested problem" as she would say. I dont understand what the parentheses mean and my trig teacher told us we wsould learn that at a latter time. Now the calc professor says we should have learned it in trig. I am getting so confused and behind. Our cakc professor teaches without using the composition of functions yet our book focusses on them so when I have a question my book is all but useless
Why don't you post a problem. What is the point of a general discussion?
Trig can be a very interesting topic. When trig is first introduce, its introduction is generally sin, cos, tan and its relationship to the right triangle. sin is opposite over hypotenuse, etc... Remember being taught that
Then we tend to teach equations with trig. You are use to 2x+3=9 well welcome to the wonderful world of trig and now your equations will be 2sin(x)+3=9 or something like that
Then eventually someone teaches the exact values of trig cos(0), sin(0), cos(pi/2) etc and the connection it has to the unit circle. but then what has become obsolete is teaching sin(22.5 degrees). With technology, you just put it in the calculator. 25 years ago, students had to use tables to determine those values.
Am I helping so far?
I pulled off an A in trig but never gained a full comprehension of it. I have to go through differential equations to start my bachelors and this has been atwo week set back and I have really studied. I am olmost contemplating quiting college because I cannot find the help I need to understand this so please I could use some clarification as to how to view trig functions. When you see cos 2x what does the 2x really represtent? our trig professor got us through the whole class without problems, or comprehension.
That is a transformation. Transformations mean up, down, right, left, vertical stretch, horizontal stretch, vertical compression, vertical stretch, reflections about the x axis or reflections about the y axis
y=2x+3 parent function is y=x +3 represents that the function moved up 3 units 2 represents a vertical stretch
In terms of cosine and sine, the transformations are called amplitude, phase shift, etc
Let me see if I can find you a website that explains it, be back in a few minutes
http://www.algebralab.org/lessons/lesson.aspx?file=Trigonometry_TrigTransformations.xml
this one has an interactive feature http://staff.argyll.epsb.ca/jreed/math30p/transformations/trigonometry.htm
Ok I remember that and think I understand it. How then does that apply to using the chain rule? and why does the use of parentheses make it so different? ex F(x)=tan^5x. My solutions guide shows the answer to be F'(x)=10 tan 5x sec^2x 5x. why do you have to apply the chain rule to it twice? And what do the answers mean?
I would not quit college. Most colleges today have math labs with tutors ready to help. Students just do not use their resources. Too many students are too busy with their electronic toys and waste alot of time. Some would not even think of going to a site like thi and talk about math. Hang in there
\[f(x)=\tan^5(x)\] Is this your problem? Please use the equation key on the lower left to post or is is \[f(x)=\tan(5x)\]
this is what wolfram alpha sates as the solution http://www.wolframalpha.com/input/?i=find+the+derivative+of+f%28x%29%3Dtan%5E5%28x%29
no it is f("theta") =tan^25"theta" there are no parentheses in the original problem
it is tangent squared five theta
how do you put it in the window like you did?
\[ f(\theta)=\tan^2(5\theta)\]
Press the blue key on the bottom, it says Equation
\[f(x)=\tan^2(5x)\] theta, x same thing
well first of all we don't have a derivative for tangent squared. But we know tan times tan is tan to the second power
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