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!!!!!
Re: Derivative of sin2x 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.
Re: Derivative of sin2x 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 ???
Re: Derivative of sin2x 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.
Re: Derivative of sin2x 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.
Re: Derivative of sin2x ^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.
Re: Derivative of sin2x 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.)