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Derivative of sin^2x 
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#1
Mar1510, 11:42 AM

P: 17

Hi.
Please tell me where this reasoning is wrong, because I know it is but I can't see how. f(x) = sinx f ' (x) = cosx f(60) = sin 60 f ' (60) = cos 60 but g(x) = sin 2x g'(x) = 2 cos 2x set x = 30 then: g(30) = sin 60 g'(x) = 2 cos 60 but f(60) = g(30) so f ' (60) = g'(30) i.e cos 60 = 2 cos 60 thus: 1 = 2 WHY!!!!! HOW!!!!! 


#2
Mar1510, 11:48 AM

P: 24

The fact that two functions cross at a particular point does not mean that their derivative is the same.
You could do the same thing with much simpler functions eg f(x)=x, g(x)=2x these cross at x=0 but they obviously don't have the same gradient. 


#3
Mar1510, 01:04 PM

P: 17

Thank you. That is quite obvious, isn't it.



#4
Mar2010, 01:19 AM

P: 83

Derivative of sin^2x
Try this one:
x=0 x+1=1 (x+1)(x+1)=1(x+1) x^2+2x+1=x+1 x^2+x=0 x(x+1)=0 x+1=0 x=1 ??? 


#5
Mar2010, 01:23 AM

P: 1,633




#6
Mar2010, 01:35 AM

P: 83

Where did it go wrong algebraically?



#7
Mar2010, 01:42 AM

P: 1,633




#8
Mar2010, 02:17 AM

PF Gold
P: 1,951

Actually, you could just as easily say that at...
x(x+1) = 0 You could just as easily divide by (x+1) to get... x = 0. After all, the first equation in this post says that x is EITHER 0 or 1. And we know that it's zero. 


#9
Mar2010, 03:02 AM

HW Helper
P: 6,202

if ab=0, then either a=0 or b=0. if you just divide by a, you'll get b=0 only. 


#10
Mar2010, 03:12 AM

P: 352

^What he said. Doing that causes a missing solution. By the end, you should get x= 1 or 0. However, 1 is an extraneous solution. Therefore, you're stuck with x=0. Check out http://en.wikipedia.org/wiki/Extraneous_solution if you're confused.



#11
Mar2010, 05:04 AM

PF Gold
P: 1,951

Yeah, I know that...
My point was that he missed a solution. 


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