How do I Find the Derivative of sin(2x)?

[SomeRandomGuy]

To find the derivative of sin(2x). So far here is what I did:

lim h -> 0 (sin(2x+2h)-sin(2x))/h
= (sin2xcos2h+cos2xsin2h-sin2x)/h
= (2sinxcosx(cos2h-1)+2sinhcoshcos2x)/h

I'm stuck here... can't remember how to simplify this. Thanks. I also left out some steps, I don't know how to use the latex graphics yet and didn't want it to get to garbled.

 

[Data]

Are you allowed to use the fact that

$$\lim_{x\rightarrow 0} \frac{\sin{\alpha x}}{\alpha x} = 1 \ \mbox{if} \ \mathbb{R} \ni \alpha \neq 0$$

? If so, it's completely trivial. Just use $$\sin{x} - \sin{a} = 2\cos{\frac{x+a}{2}} \sin{\frac{x-a}{2}}$$.

 

[robphy]

If you know
lim h -> 0, (sin(x+h)-sin(x))/h = cos(x),
then

lim h -> 0,
(sin(2x+2h)-sin(2x))/h
= (2/2)*(sin(2x+2h)-sin(2x))/h ... that is, multiply by 1 in a funny way
= 2(sin(2x+2h)-sin(2x))/(2h)
= 2(sin(X+H)-sin(X))/(H) where X=2x and H=2h (note that lim h->0 implies lim H->0)

= 2cos(X)
= 2cos(2x)

 

[Data]

Haha, somehow I got the idea that you wanted to find the derivative of $$\sin{x}$$ (which is actually "harder"). All you need for $$\sin{2x}$$ is

$$\lim_{x\rightarrow 0} \frac{\sin{x}}{x} = 1$$