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differentiate sin^2x |
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| May7-07, 08:56 AM | #1 |
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differentiate sin^2x
How do you differentiate the likes of (sinx)^2
thanks |
| May7-07, 08:58 AM | #2 |
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Mentor
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Use the chain rule. Let u=sinx, then you need to find d/dx(u^2).
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| May7-07, 09:05 AM | #3 |
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Alternatively, you can recall / derive the power reduction formulae such as;
[tex]\sin^2\theta = \frac{1 - \cos 2\theta}{2}[/tex] These are especially useful when integrating such functions. |
| May7-07, 10:17 AM | #4 |
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Recognitions:
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differentiate sin^2x
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| May7-07, 05:06 PM | #5 |
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Recognitions:
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I Think Hootenanny was in fact intended to use that to simplify the differentiation, if I'm reading his last sentence correctly >.<...Well anyway It doesn't really help very much because we still have to use the chain rule on the cos 2theta.
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| May8-07, 03:33 AM | #6 |
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[tex]\frac{d}{dx}\sin(ax) dx = a\cos(ax)[/tex] [tex]\int \sin(ax) dx = -\frac{1}{a}\cos(ax) + C[/tex] Rather than remembering the results for the sin2x etc. In any event applying the chain rule to something of the form sin(ax) is somewhat simpler than applying it to something of the form sin2x don't you think? |
| May8-07, 03:39 AM | #7 |
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Not always, usaully one would like an answer without double angled arguments, so they would have to know the expansion of cos(2theta) which isn't as easy as bringing a power down times the derivative of sin x.
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| May8-07, 03:47 AM | #8 |
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| May8-07, 04:29 AM | #9 |
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Of course there all equivalent, but I always prefer putting my answers in terms in single angled arguments. In the end it makes very little difference, maybe 5 seconds working time.
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| May8-07, 10:50 AM | #10 |
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= 2sinx cosx
= sin2x
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| May9-07, 05:59 PM | #11 |
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w00t 1st post :P |
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