# Derivative of sin^2x

by Nerd
Tags: derivative, sin2x
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 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!!!!!
 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.
 P: 17 Thank you. That is quite obvious, isn't it.
 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 ???
P: 1,633
 Quote by darkside00 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!
 P: 83 Where did it go wrong algebraically?
P: 1,633
 Quote by darkside00 Where did it go wrong algebraically?
I just told you! I boldfaced those parts.
 PF Gold P: 1,957 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.
HW Helper
P: 6,202
 Quote by Char. Limit 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.
 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.
 PF Gold P: 1,957 Yeah, I know that... My point was that he missed a solution.
 Mentor P: 12,074 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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