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Find the derivative and the domain of the derivative (trig funtions) 
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#1
Nov1011, 01:04 PM

P: 28

1. The problem statement, all variables and given/known data
f(x)=sin(2x+5) 2. Relevant equations I am supposed to find the derivative, and the domain of it. The problem is that I cannot deal with the simplification of it. 3. The attempt at a solution As far as I know, I am suppsosed to use f(x+h)f(x)/h = sin(2(x+h)+5)(sin(2x+5))/h I tried expanding what's inside, as in sin(2x+2h+5)sin(2x+5)/h But that doesn't seem to lead to a solution 


#2
Nov1011, 01:23 PM

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#3
Nov1011, 01:28 PM

P: 28

I have learned the chain rule, and I am certain I am supposed to use the limit definition in this question



#4
Nov1011, 02:08 PM

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P: 21,216

Find the derivative and the domain of the derivative (trig funtions)
OK, then I think you'll need to use the sum identity for sine: sin(A + B) = sin(A)cos(B) + cos(A)sin(B).
You will probably need to use a couple of limit formulas as well: [tex]\lim_{h \to 0}\frac{sin(h)}{h} = 1[/tex] [tex]\lim_{h \to 0}\frac{cos(h)  1}{h} = 0[/tex] 


#5
Nov1011, 03:50 PM

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P: 2,263

recall the trigonometric identity
sin(2(x+h)+5)(sin(2x+5))=2 sin(h) cos(h+2 x+5) or in general sin(A)sin(B)=2 sin(A/2B/2) cos(A/2+B/2) and use the facts that cosine is continuous sin'(0)=1 


#6
Nov1111, 02:38 PM

P: 28

Am I supposed to solve for when a > 0? That's what I did. sin(2x+5)sin(2a+5) = 2sin ((xa)/2) cos ((x+a)/2) = sin (2x+5)sin (5) = 2sin (x/2) cos (x/2) Then the x was = 0 I am not sure I am on the right track. The derivative is cos(2x+5)(2) .. I am not supposed to achieve a value. So was I supposed to keep the a? 


#7
Nov1111, 04:32 PM

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P: 7,795

You need to look at sin((2x+5)+h)sin(2a+5) = 2sin (h/2) cos ((2x+5)+h/2) . 


#8
Nov1111, 04:38 PM

P: 28

Oh. Alright. Thank you very much :] 


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