# Find the derivative and the domain of the derivative (trig funtions)

by Absolutism
Tags: derivative, domain, funtions, trig
 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
Mentor
P: 21,215
 Quote by Absolutism 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
Are you certain that you have to use the limit definition of the derivative here? Have you learned the chain rule yet?
 P: 28 I have learned the chain rule, and I am certain I am supposed to use the limit definition in this question
 Mentor P: 21,215 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: $$\lim_{h \to 0}\frac{sin(h)}{h} = 1$$ $$\lim_{h \to 0}\frac{cos(h) - 1}{h} = 0$$
 HW Helper 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/2-B/2) cos(A/2+B/2) and use the facts that cosine is continuous sin'(0)=1
P: 28
 Quote by Absolutism I have learned the chain rule, and I am certain I am supposed to use the limit definition in this question
 Quote by Mark44 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: $$\lim_{h \to 0}\frac{sin(h)}{h} = 1$$ $$\lim_{h \to 0}\frac{cos(h) - 1}{h} = 0$$
 Quote by lurflurf 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/2-B/2) cos(A/2+B/2) and use the facts that cosine is continuous sin'(0)=1

Am I supposed to solve for when a --> 0?

That's what I did.

sin(2x+5)-sin(2a+5) = 2sin ((x-a)/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?
Emeritus
HW Helper
PF Gold
P: 7,800
 Quote by Absolutism Am I supposed to solve for when a --> 0? That's what I did. sin(2x+5)-sin(2a+5) = 2sin ((x-a)/2) cos ((x+a)/2) ...
Actually sin(2x+5)-sin(2a+5) = 0

You need to look at sin((2x+5)+h)-sin(2a+5) = 2sin (h/2) cos ((2x+5)+h/2) .
P: 28
 Quote by SammyS Actually sin(2x+5)-sin(2a+5) = 0 You need to look at sin((2x+5)+h)-sin(2a+5) = 2sin (h/2) cos ((2x+5)+h/2) .

Oh. Alright. Thank you very much :]

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