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

[Absolutism]

Homework Statement

f(x)=sin(2x+5)

Homework 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.

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

 

[Mark44]

Homework Statement

f(x)=sin(2x+5)

Homework 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.

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]

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

 

[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

 

[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

 

[Absolutism]

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

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

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]

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]

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 :]