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

by Absolutism
Tags: derivative, domain, funtions, trig
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Absolutism
#1
Nov10-11, 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
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Mark44
#2
Nov10-11, 01:23 PM
Mentor
P: 21,413
Quote Quote by Absolutism View Post
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?
Absolutism
#3
Nov10-11, 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

Mark44
#4
Nov10-11, 02:08 PM
Mentor
P: 21,413
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]
lurflurf
#5
Nov10-11, 03:50 PM
HW Helper
P: 2,264
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
Absolutism
#6
Nov11-11, 02:38 PM
P: 28
Quote Quote by Absolutism View Post
I have learned the chain rule, and I am certain I am supposed to use the limit definition in this question
Quote Quote by Mark44 View Post
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]
Quote Quote by lurflurf View Post
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?
SammyS
#7
Nov11-11, 04:32 PM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
Quote Quote by Absolutism View Post
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) .
Absolutism
#8
Nov11-11, 04:38 PM
P: 28
Quote Quote by SammyS View Post
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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