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Use definition of derivative |
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| Mar15-05, 09:25 PM | #1 |
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Use definition of derivative
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. |
| Mar15-05, 09:51 PM | #2 |
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Are you allowed to use the fact that
[tex] \lim_{x\rightarrow 0} \frac{\sin{\alpha x}}{\alpha x} = 1 \ \mbox{if} \ \mathbb{R} \ni \alpha \neq 0[/tex] ? If so, it's completely trivial. Just use [tex]\sin{x} - \sin{a} = 2\cos{\frac{x+a}{2}} \sin{\frac{x-a}{2}}[/tex]. |
| Mar15-05, 10:55 PM | #3 |
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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) |
| Mar15-05, 11:01 PM | #4 |
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Use definition of derivative
Haha, somehow I got the idea that you wanted to find the derivative of [tex]\sin{x}[/tex] (which is actually "harder"). All you need for [tex]\sin{2x}[/tex] is
[tex]\lim_{x\rightarrow 0} \frac{\sin{x}}{x} = 1[/tex] |
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