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Differentiate sin^2x

by strokebow
Tags: differentiate, sin2x
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strokebow
#1
May7-07, 08:56 AM
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How do you differentiate the likes of (sinx)^2

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cristo
#2
May7-07, 08:58 AM
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Use the chain rule. Let u=sinx, then you need to find d/dx(u^2).
Hootenanny
#3
May7-07, 09:05 AM
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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.

Curious3141
#4
May7-07, 10:17 AM
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Differentiate sin^2x

Quote Quote by Hootenanny View Post
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.
Differentiating, not integrating.
Gib Z
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May7-07, 05:06 PM
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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.
Hootenanny
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May8-07, 03:33 AM
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Quote Quote by Gib Z View Post
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.
Not at all, I meant that the formulae are useful for differentiation, but more so for integration. It is true that to differentiate you may have to use the chain rule for both forms but I find it easier to remember that;

[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?
Gib Z
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May8-07, 03:39 AM
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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.
Hootenanny
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May8-07, 03:47 AM
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Quote Quote by Gib Z View Post
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.
Fair point perhaps, but I've never come across a case where a single angle argument is preferable to a double angle. In any case, the double angle form is certainly much easier to integrate.
Gib Z
#9
May8-07, 04:29 AM
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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.
DAKONG
#10
May8-07, 10:50 AM
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= 2sinx cosx
= sin2x
fblade
#11
May9-07, 05:59 PM
P: 2
Quote Quote by DAKONG View Post
= 2sinx cosx
= sin2x
ditto..


w00t 1st post :P
Office_Shredder
#12
May9-07, 06:08 PM
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Let's try not to get into a 12 page debate on the merits of single vs. double angle final form solutions


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