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Derivative of sin^2x

  • Thread starter Nerd
  • Start date
  • #1
17
0
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!!!!!
 
Last edited by a moderator:

Answers and Replies

  • #2
29
4


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
17
0


Thank you. That is quite obvious, isn't it.
 
  • #4
83
0


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
1,631
4


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

???
:wink: Ops!
 
  • #6
83
0


Where did it go wrong algebraically?
 
  • #7
1,631
4


Where did it go wrong algebraically?
I just told you! I boldfaced those parts.
 
  • #8
Char. Limit
Gold Member
1,204
13


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
rock.freak667
Homework Helper
6,230
31


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.
 
  • #10
354
2


^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
Char. Limit
Gold Member
1,204
13


Yeah, I know that...

My point was that he missed a solution.
 
  • #12
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129


Moderator's note:

Please keep on topic with the thread. New topics should be started in a new thread.

(See post #1 if you don't know what the topic of this thread is.)
 

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