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Differentiate sin^2x 
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#1
May707, 08:56 AM

P: 101

How do you differentiate the likes of (sinx)^2
thanks 


#2
May707, 08:58 AM

Mentor
P: 8,316

Use the chain rule. Let u=sinx, then you need to find d/dx(u^2).



#3
May707, 09:05 AM

Emeritus
Sci Advisor
PF Gold
P: 9,772

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. 


#5
May707, 05:06 PM

HW Helper
P: 3,348

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.



#6
May807, 03:33 AM

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PF Gold
P: 9,772

[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 sin^{2}x 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 sin^{2}x don't you think? 


#7
May807, 03:39 AM

HW Helper
P: 3,348

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.



#8
May807, 03:47 AM

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P: 9,772




#9
May807, 04:29 AM

HW Helper
P: 3,348

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.



#10
May807, 10:50 AM

P: 15

= 2sinx cosx
= sin2x 


#11
May907, 05:59 PM

P: 2

w00t 1st post :P 


#12
May907, 06:08 PM

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P: 4,500

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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