derivative of sin^2x


by Nerd
Tags: derivative, sin2x
Nerd
Nerd is offline
#1
Mar15-10, 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!!!!!
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cpt_carrot
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#2
Mar15-10, 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.
Nerd
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#3
Mar15-10, 01:04 PM
P: 17
Thank you. That is quite obvious, isn't it.

darkside00
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#4
Mar20-10, 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

???
sutupidmath
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#5
Mar20-10, 01:23 AM
P: 1,635
Quote Quote by darkside00 View Post
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

???
Ops!
darkside00
darkside00 is offline
#6
Mar20-10, 01:35 AM
P: 83
Where did it go wrong algebraically?
sutupidmath
sutupidmath is offline
#7
Mar20-10, 01:42 AM
P: 1,635
Quote Quote by darkside00 View Post
Where did it go wrong algebraically?
I just told you! I boldfaced those parts.
Char. Limit
Char. Limit is offline
#8
Mar20-10, 02:17 AM
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P: 1,930
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.
rock.freak667
rock.freak667 is offline
#9
Mar20-10, 03:02 AM
HW Helper
P: 6,212
Quote Quote by Char. Limit View Post
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.
If you do that you might miss out a solution.

if ab=0, then either a=0 or b=0.

if you just divide by a, you'll get b=0 only.
Anonymous217
Anonymous217 is offline
#10
Mar20-10, 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.
Char. Limit
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#11
Mar20-10, 05:04 AM
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P: 1,930
Yeah, I know that...

My point was that he missed a solution.
Redbelly98
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#12
Mar20-10, 09:05 AM
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