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March 10th, 2013, 01:18 PM  #1 
Newbie Joined: Mar 2013 Posts: 4 Thanks: 0  what is the differential of a trigonometric function
I have a calculus question I can't figure out how to solve, The function y = tan(3x + 6); what is the dy when x=3 and dx=0.03 I tried these steps: f'(x)= sec^2(3x+6)(3)dx = (1/cos^2(3x+6))(3)dx, (need to calculate in radians) so, for f'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) f'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) = (1/[(1+cos2(15))/2]) = (2/(1+cos30)) = {2/[1+ sqrt(3)/2]} = [2/[(2+sqrt(3))/2]] = 4/((2+sqrt(3)), so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3)) = and would be ..... am I doing this right? Thank you. 
March 10th, 2013, 03:15 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,931 Thanks: 2205 
You are incorrect. You mentioned "need to calculate in radians", but didn't clarify whether x is already in radians. When you evaluated cos(30), you seemed to be working in degrees, but that would mean that everything was already in degrees and should have been converted to radians from the outset to make the differentiation valid.

March 10th, 2013, 03:50 PM  #3 
Newbie Joined: Mar 2013 Posts: 4 Thanks: 0  Re: what is the differential of a trigonometric function
My mistake, yes worked as everything is in degrees, what I mentioned about radians it's because of the transformations I needed to do in order to use the calculator, what would you think about if you see the problem in this way: The function y = tan(3x + 6); what is the dy when x=3 and dx=0.03 I tried these steps: f'(x)= sec^2(3x+6)(3)dx = (1/cos^2(3x+6))(3)dx, so, for f'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) f'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) = (1/[(1+cos2(15))/2]) = (2/(1+cos30)) = {2/[1+ sqrt(3)/2]} = [2/[(2+sqrt(3))/2]] = 4/((2+sqrt(3)), so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3)) = 
March 10th, 2013, 04:39 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,931 Thanks: 2205 
Why have you repeated three lines for no apparent reason? The trigonometric manipulation is valid, but the calculus isn't. You can't use sec²(x) as the derivative of tan(x) unless x is in radians.

March 10th, 2013, 04:45 PM  #5 
Newbie Joined: Mar 2013 Posts: 4 Thanks: 0  Re: what is the differential of a trigonometric function
so I have to transform y = tan(3x + 6) in radians and after that to obtain the derivative, or I just have to assume that it was from the beginning in radians? sorry, but the problem doesn't state whether x is in radians or in degrees and I assumed from the beginning that it is in degrees, I'm completely confused now?

March 11th, 2013, 04:35 PM  #6 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: what is the differential of a trigonometric function ALL derivatives of trig functions are based on radians (Technically, the trig functions can be defined without reference to angles at all. "Radian" measure for angles happens to give the same derivatives.) Unless a problem specifically refers to angle measured in degrees, 'radians' are to be assumed.

March 11th, 2013, 05:05 PM  #7 
Newbie Joined: Mar 2013 Posts: 4 Thanks: 0  Re: what is the differential of a trigonometric function
So, am I finally right in my calculations or not, and if not can you show where is my mistake f'(x)= sec^2(3x+6)(3)dx = (1/cos^2(3x+6))(3)dx, for f'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) = (1/[(1+cos2(15))/2]) = (2/(1+cos30)) = {2/[1+ sqrt(3)/2]} = [2/[(2+sqrt(3))/2]] = 4/((2+sqrt(3)), so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3))... 
March 13th, 2013, 09:29 AM  #8  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: what is the differential of a trigonometric function Quote:
 

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