# Derivative of sin^2x

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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## Answers and Replies

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.

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

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

???

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!

Where did it go wrong algebraically?

Where did it go wrong algebraically?

I just told you! I boldfaced those parts.

Char. Limit
Gold Member

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
Homework Helper

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.

^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
Gold Member

Yeah, I know that...

My point was that he missed a solution.

Redbelly98
Staff Emeritus
Science Advisor
Homework Helper

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